Calculus Problem Set 2-1-37 Ge 1a

1. What number exceeds its square by the maximum amount? (Desuasido, Jhenelyn) Let x = the number and x2 = the square of the number y = the difference between x and x2 y = x−x2 y′ = 1−2x = 0 y′ = 1−2x = 0 1 𝑥=2 Answer 𝟏

Therefore, the number that exceeds its square by the maximum amount is 𝟐. 2. What positive number added to its reciprocal gives the minimum sum? (Durian, Mary Louisse) x = positive number ( x>0) 1 𝑥

= reciprocal of x

y = sum of x and 1/x Solution: 1. Make an equation out of the given variables. 𝒚=𝒙+

𝟏 𝒙 1

2. Get the derivative of 𝑦 = 𝑥 + 𝑥 . 𝑦 ′ = 1 + 𝑥 −1 𝑦 ′ = 1 + 𝑥 −1−1 𝑦 ′ = 1 − 𝑥 −2 𝟏

𝒚 ′ = 𝟏 − 𝒙𝟐 3. To test the critical point, get the second derivative. 𝑦 ′ = 1 − 𝑥 −2 𝑦 ′′ = 2𝑥 −2−1 𝑦 ′′ = 2𝑥 −3 𝒚′′ =

𝟐 𝒙𝟑

4. Get the critical points. Let: 𝑦 ′ = 0 𝑦′ = 1 − 1−

1 𝑥2

1 =0 𝑥2

𝑥2 − 1 =0 𝑥2 𝑥2 − 1 = 0 (𝑥 − 1)(𝑥 + 1) = 0 𝑥=1 𝒙 = −𝟏 5. Using the second derivative, test the critical points. Get the point that satisfies the question. 𝑦 ′′ =

2 𝑥3

Substitute: if x = 1 𝑦 ′′ =

2 =2 13

If x = -1 𝒚′′ =

𝟐 = −𝟐 −𝟏𝟑

3. ) The sum of two numbers is k. Find the minimum value of the sum of their squares. (Divina, Mark Ryan) EQUATION 1:

x+ y =k

let x and y be the two numbers and k as their sum.

y=k−x

find y

EQUATION 2:

x +y 2

2

=s

let s be the sum of their sqaures

s = x 2 + (k − x) 2

substitute equation 1 into equation 2

ds = x 2 + (k − x) 2 dx

find the derivative of s with respect to x

= 2 x + 2(k − x)(−1)

= 2x + (−2k + 2 x) = 2 x − 2k + 2 x

use the chain rule for (k − x) 2 distribute -2 get rid of the parenthesis

= 4x − 2k

combine like terms

4x 2 = k 4 4

move − 2k to the other side of the equation

x=

1 k 2

1 y=k− k 2

subtitute the value of x to the first equation

and divide both sides by 4

y=

1 k 2

s = (1 2 k ) 2 + (1 2 k ) 2 substitute the value of x and y to the second equation =

1 2 1 2 k + k 4 4

s=

1 2 k 2

4. The sum of two numbers is k. Find the minimum value of the sum of their cubes. (Boongaling, Rheanell) Solution: Let, x and y = the numbers z = sum of their cubes

𝑘 =𝑥+𝑦 𝑦 =𝑘−𝑥 𝑧 = 𝑥³ + 𝑦³ 𝑧 = 𝑥² + (𝑘 − 𝑥)³ 𝑑𝑧 = 3𝑥² + 3(𝑘 − 𝑥)²(−1) = 0 𝑑𝑥 𝑥² − (𝑘² − 2𝑘𝑥 + 𝑥²) = 0 𝑥² − 𝑥 2 − 𝑘 2 − 2𝑘𝑥 = 0 −𝑘 2 − 2𝑘𝑥 = 0 2𝑘𝑥 𝑘 2 = 2𝑘 2𝑘 1

𝑥 = 2𝑘

1

𝑦 = 𝑘 − 2𝑘 1

𝑦 = 2𝑘 1

1

𝑧 = (2 𝑘)3 + (2 𝑘)3 1 1 𝑧 = 𝑘3 + 𝑘3 8 8 𝒛=

𝟏 𝒌³ 𝟒

5. The sum of 2 positive numbers is 2. Find the smallest value possible for the sum of the cube of one number and the square of the other? (Espinola, Christian) Let x and y be the 2 positive numbers 𝑥+𝑦 =2 1 + 𝑦′ = 0 𝑦 ′ = −1 𝑧 = 𝑥3 + 𝑦2 𝑑𝑧 = 3𝑥 2 𝑑𝑥

+ 2𝑦𝑦′ = 0

3𝑥 2 + 2𝑦(−1) = 0 3𝑥 2 − 2𝑦 = 0 3𝑥 2 =2𝑦 2

𝑦=

3 2

𝑥2

𝑥+𝑦 =2 2 [𝑥 +

3 2 𝑥 = 2] 2 2

2𝑥 + 3𝑥 2 = 4 3𝑥 2 + 2𝑥 − 4 = 0 a = 3, b = 2, c = -4 = = = =

−𝑏±√(𝑏)2 + 4𝑎𝑐 2𝑎

−2 ± √(2)2 + 4(3)(−4) 2(3) −2±√4+48

=

6 −2+2√13

=

6

𝑥 = 0.8685

3

𝑥2 2 3

𝑦 = 2 (0.8685)2 𝑦 = 1.1314

6

−2−2√13 6

& 𝑥 = −1.5352

𝑥 = 0.8685 𝑦=

−2±2√13

𝑥 = −1.5352 3

𝑦 = 2 𝑥2 𝑦=

3 2

(−1.5352)2

𝑦 = 3.5353 - not possible

𝑧 = 𝑥3 + 𝑦2 𝑧 = (0.8685)3 + (1.1314)2 𝑧 = 0.6551 + 1.2801 𝒛 = 𝟏. 𝟗𝟑𝟓𝟐

6.Find two numbers whose sum is a, if the product of one by the square of the other is to be a maximum. (Estrellado, Marlon) x+y=a u=xy² u=(a-y)y² u=ay²-y³ du/dy=2ay-3y² 2ay-3y²=0 y=(2/3)a x=a-y =a-(2/3)a x=(1/3)a

7. Find two numbers whose sum is a, if the product of one by the cube of the other is to be a maximum. (Dichoso, Rosely)

Solution: Let x and y the numbers x+y=a x=a−y z=xy3 z=(a−y)y3 z=ay3−y4 dz/dy=3ay2−4y3=0 y2(3a−4y)=0 3 y=0(absurd) and 4a(use) 3

x=a−4a 1

x=4a 𝟏

𝟑

The numbers are: 𝟒 𝒂 and 𝟒 𝒂.

8. Find two numbers whose sum is a, if the product of the square of one by the cube of the other is to be a maixum.(Divina, Keanne) Let b and c be the numbers. b+c=a Get the derivative in terms of c. 1+c′ =0 c′ = −1 Let j be he product of the square of one by the cube of the other. j= b2c3 𝑑𝑗 𝑑𝑏

= 𝑏 2 (3𝑐 2 𝑐 ′ ) + (2𝑏)𝑐 3

𝑏 2 (3𝑐 2 𝑐 ′ ) + (2𝑏)𝑐 3 = 0 𝑏𝑐 2 [𝑏 2 (3𝑐 2 𝑐 ′ ) + (2𝑏)𝑐 3 ] = 0 3𝑏𝑐′ + 2𝑐 = 0 3𝑏(−1) + 2𝑐 = 0 −3𝑏 = −2𝑐 𝟐

𝒃 = 𝟑𝒄

2

𝑏+𝑐 =𝑎 2 3

𝑏 = 3𝑐 2 3

𝑐+𝑐 =𝑎

𝑏 = 3 (5 𝑎)

5

𝟐

𝑐=𝑎 3

𝒃 = 𝟓𝒂

𝟑

𝒄 = 𝟓𝒂 𝟐

𝟑

Therefore the two numbers are 𝟓 𝒂 and 𝟓 𝒂.

9. What should be the shape of a rectangular field of a given area, if it is to be enclosed by the least amount of fencing? (Dacayo, Jimnuel) Area: A= xy 0=xy′+ y y′=− y/x Perimeter: P=2x+2y

dP dx

=2+2y′=0 −y

1+ ( ) = 0 x

y=x (a square)

10. A rectangular field of given area is to be fenced off along the bank of a river. If no fence is needed along the river, what is the shape of the rectangle requiring the least amount of fencing? (Dipon, Kenneth) Illustration:

y

x •

The formula for finding the area of the rectangle is; 𝐴=𝐿∗𝑊

•

By substituting the variables, we can get; 𝐴 = 𝑥𝑦

•

Derive and equate it to zero 𝑥𝑦’ + 𝑦 = 0

•

Equate the equation to 𝑦’ 𝑦′ = −

•

𝑦 𝑥

And the formula for finding the perimeter of a rectangle is; 𝑃 = 2𝐿 + 2𝑊 𝑃 = 2𝑥 + 2𝑦

•

Since the other side of the field doesn’t need to be fenced, then the formula for the perimeter of the field is 𝑃 =𝑥+2

•

By deriving it we will get; 𝑑𝑃 = 1 + 2𝑦 ′ = 0 𝑑𝑥

•

Then substitute the value of 𝑦′;

𝑑𝑃 𝑦 = 1 + 2 (− ) = 0 𝑑𝑥 𝑥 •

And we will get; 𝒚=

𝟏 𝒙 𝟐

So the shape of the rectangle requiring the least amount of fencing is ½*length 11. A rectangular lot is to be fenced off alonga highway. If the fence on the highway costs m dollars per yd, on the other sides n dollars per yd, find the area of the largest lot that can be fenced off for k dollars. (Deocareza, Neil Jhon) Let x=width, y=length The total cost will be: k=mx+n(2y+x) because only one side x costs m dollars while the rest of the lengths cost n. Simplifying: k = (m + n)x + 2ny 2ny = k − (m + n)x First equation

y=

k − (m + n)x 2n

The Area is A = xy Substituting the first equation: 𝐴 = 𝑥(

k − (m + n)x ) 2n

xk − (m + n)x 2 𝐴= 2n Differentiating: 𝑑𝐴 1 2(m + n)x = ( )k − 𝑑𝑥 2𝑛 2n (m + n)x 𝑑𝐴 1 = ( )k − 𝑑𝑥 2𝑛 n Equating to 0 to get the critical x value and maxima: (m + n)x 1 ( )k − =0 2𝑛 n 𝑘 − 2(𝑚 + 𝑛)𝑥 = 0 𝑥=

𝑘 2(𝑚 + 𝑛)

Substituting into the first equation: y=

k−(m+n)( 2n

𝑘 ) 2(𝑚+𝑛)

=

𝑘 2

k−( ) 2n

=

2k−k 2

2n

k

= 4n

Solving for the maximum area:

A= =

𝑘

𝑘

( )

2(𝑚+𝑛) 4𝑛 𝒌𝟐

𝟖𝒏(𝒎+𝒏)

𝒚𝒅𝟐

12. A rectangular field of fixed area is to be enclosed and divided into three lots by parallels to one of the sides. What should be the relative dimensions of the field to make the amount of fencing minimum? (Atok, Mark) x

y

Using the formula of the area: Area:L x W A=xy Get the derivative of the area: 0=xy′+y y′=−y/x Relate the dx/dy of the area to the 1st derivative of the perimeter of the rectangle: Fence: P=2x+4y dP/dx=2+4y′=0 2+4(−y/x)=0 y=1/2x width = ½ × length 13. Do example 12 with the words “three lots” replaced by “five lots.” (Fajardo, Jeselle)

You need to use the formula in getting the area of the rectangle. Let base be x and height be y. Get the derivative of the formula for area (which is in implicit form).

X

𝐴 = 𝑏ℎ 𝐴 = 𝑥𝑦 0 = 𝑥𝑦 ′ + y 𝑦 ′ = −𝑦/x

Y

Fence: 𝑃 = 2𝑥 + 6𝑦 𝑑𝑃 𝑑𝑥

= 2𝑥 + 6𝑦 ′ = 0

Use the formula for the parameter of rectangle and find its derivative 𝑦 to+get 2 6 the (− length ) = 0of the width of the field. x 𝑦=

1 𝑥 3

The derivative of the parameter is 1/3 x.

1

Width= 3 𝑥 lenght

14. A rectangular lot is bounded at the back by a river. No fence is needed along the river and there is to be a 24 ft. opening in front. If the fence along the front costs $1.5 per ft., along the sides$1 per ft., find the dimensions of the largest lot which can be thus fenced in for $300. (Dometita, Ramona) Solution: Let L= length and W= width cost of fence for L= (L-24)(1.5)= 1.5 L - 36 cost of fence for W= (W+W)(1)= 2W (1.5L - 36)+2W= 300 1.5L + 2W= 336 W= 336 - 1.5L(1/2)

Area of rectangle= LW

A= L(336 - 1.5L)(1/2)

A= 336L - 1.5L^2(1/2) dA/dL= 1/2(336 - 1.5L) d^2A/dL^2= -3/4 dA/dL= 0 1/2(336 - 1.5L)=0

L= 112 ft.

Then we can compute for width: W= 336 -1.5L(1/2) W= 336 - 1.5(112)(1/2)

Answer: 112 ft. by 84 ft.

W= 84 ft.

15. A box is to be made of a piece of cardboard 9 inches square by cutting equal squares out of the corners and turning up the sides. Find the volume of the largest box that can be made in this way. (Estoperez, Renalyn)

*h = x

* l × w = (9 - 2x)2 V = l ×w×h

V = (9 − 2 x)2 x *Expand

V = (81 − 36x + 4 x2 ) x

V = 81x − 36x2 + 4 x3

dV = 12 x 2 − 72 x + 81 dx *Divide all terms by 3

4 x2 − 24x + 27 = 0 *Use quadratic formula:

x= A=4

− b  b 2 − 4ac 2a b=-24

c=27

x=

− (−24)  (−24) 2 − 4(4)(27) 2(4)

x=

24  144 8

x=

24  12 8

x1=4.5 is too high, *Use x2=1.5, then substitute:

V = (9 − 2 x)2 x

V = [(9 − 2(1.5in)]2 (1.5in)

V = 54in3

16. Find the volume of the largest box that can be made by cutting equal squares out of the corners of a piece of cardboard of dimension 15 in. by 24 in., and then turning up the sides. (Flores, Dinalyn)

15 inches 𝑥 24 inches

(15 − 2𝑥)

(24 − 2𝑥)

Formula for Volume: •

𝑉 = 𝐿𝐻𝑊

Let x be the side that will be cut to the rectangle. •

𝑉 = 𝑥(15 − 2𝑥)(24 − 2𝑥)

Get the product, • •

= (15𝑥 − 2𝑥 2 )(24 −2x) = 360𝑥 −78𝑥 2 + 4𝑥 3

(4𝑥 3 − 78𝑥 2 + 360𝑥)

Find the derivative of the equation: •

We have 2 value for x, which are x=10 and x=3, then substitute the value in the original equation:

𝑉 = 4𝑥 3 − 78𝑥 2 + 360

𝑉′ = 4𝑥 3 − 78𝑥 2 + 360 𝑉′ = (3(4𝑥)

3−1

) − (2(78𝑥)

1. x=10 2−1

) + 360(1)

𝑉 = 10(15 − 2(10))(24 − 2(10))

2

𝑉′ = 12𝑥 − 156𝑥 + 360 𝑉′ =

12𝑥 2 −156𝑥+360 12

𝑉 ′ = 𝑥 2 − 13𝑥 + 30 𝑉 ′ = (𝑥 − 10)(𝑥 − 3)

= -200 *The 𝑥 = 10 is not valid because no area is in (-) negative. 2. x=3 𝑉 = 3(15 − 2(3))(24 − 2(3))

𝑥 − 10 = 0 𝑥 = 10

𝑥−3 =0 𝑥=3

=486 𝒊𝒏.𝟑 The volume of a largest box is 486 𝒊𝒏.𝟑

17. Find the depth of the largest box that can be made by cutting equal squares of side x out of the corners of a piece of cardboard of dimensions 6a, 6b, (b ≤ a), and then turning up the sides. To select that value of x which yields a maximum volume, show that

(a+b+a^2−√{ab+b^2}) ≥3b (Deramas, Gary)

V=(6a−2x)(6b−2x)x V=36abx−12(a+b)x^2+4x^3 dV/dx=36ab−24(a+b)x+12x^2=0 x^2−2(a+b)x+3ab=0 A=1;B=−2(a+b);C=3ab x=−B±√{B2−4AC}/2A x=2(a+b)± √{4(a+b)2−4(1)(3ab)}/2(1) x=2(a+b)±2√ {(a^2+2ab+b^2)−3ab}/2 x=(a+b)+ √{a^2−ab+b^2} and

x=(a+b)− √{a2−ab+b2} If a = b: From

x=(a+b)+ √{a2−ab+b2} x=(b+b)+ √{b2−b2+b2} x=3b From

x=(a+b)− √{a^2−ab+b^2} x=(b+b)− √{b^2−b^2+b^2} x=b Answer Use x=a+b−√{a^2−ab+b^2}

18. The strength of a rectangular beam is proportional to the breadth and the square of the depth. Find the shape of the strongest beam that can be cut from a log of given size. (Ebrada, Aevy) Let s=strength, b=breadth, d=depth and D=diameter of the log. The illustration of this problem is D

d

b

The diameter 𝐷 of the log is given, thus 𝐷 is constant. Using the Pythagorean Theorem, 𝑏2 + 𝑑2 = 𝐷2 Using the equation we should find the derivative of breadth with respect to depth

𝑑𝑏 𝑑𝑑

𝑏2 + 𝑑2 = 𝐷2 𝑑𝑏

𝑑𝑏

(The derivative of 𝑏 2 is 2𝑏 𝑑𝑑 , the derivative of 𝑑 2 is 2𝑑 and the

2𝑏 𝑑𝑑 + 2𝑑 = 0

derivative of the constant 𝐷2 is 0.) 𝑑𝑏

2𝑏 𝑑𝑑 = −2𝑑 Dividing both sides by 2𝑏, 𝑑𝑏 −𝑑 = 𝑑𝑑 𝑏

Based on the problem given, 𝑠 = 𝑏𝑑2 𝑑𝑠

Finding the 𝑑𝑑 , we would use the product rule since the equation for strength is given by the product between breadth 𝑏 and depth 𝑑. 𝑑𝑠 𝑑𝑑

𝑑𝑏

= 𝑏(2𝑑) + 𝑑 2 𝑑𝑑 𝑑𝑠

Since this is a problem concerning maxima, then 𝑑𝑑 = 0. 𝑏(2𝑑) + 𝑑 2

𝑑𝑏 =0 𝑑𝑑

𝑑𝑏

Substituting the value of 𝑑𝑑 which is 𝑏(2𝑑) + (𝑑 2 )

−𝑑 𝑏

,

−𝑑 =0 𝑏

2𝑏𝑑 −

𝑑3 =0 𝑏

2𝑏𝑑 =

𝑑3 𝑏

Multiplying both sides by 𝑏 and dividing by 𝑑, 2𝑏 2 = 𝑑 2 𝑑 = √2𝑏 2 𝒅 = 𝒃√𝟐 Therefore, the depth of the strongest beam that can be cut from a log of given size is the product of √2 and breadth.

19. The stiffness of a rectangular beam is proportional to the breadth and the cube of the depth. Find the shape of the stiffest beam that can be cut from a log of given size. (Fordelon, Bryan) Diameter is given (log of given size), thus D is constant 𝑏 2 +𝑑2 =𝐷2

𝑑𝑏

2b 𝑑𝑑 + 2𝑑 = 0

𝑑𝑏 𝑑 =− 𝑑𝑑 𝑏

Stiffness: k = 𝑏𝑑 3 𝑑𝑘 𝑑𝑑

𝑑𝑏

= 𝑏(3𝑑 2 ) + 𝑑 3 𝑑𝑑 = 0 𝑑

3𝑏𝑑2 +𝑑 3 (− 𝑏 )=0 3𝑏 2 = 𝑑 3 3𝑏 2 =𝑑2

𝑑4 𝑏

d=√3𝑏 Depth =√𝟑 x breadth

20. Compare for strength the stiffness, against both edgewise and sidewise thrust, two beams of equal length, one 2 in. by 8 in. the other 4 in. by 6 in. ( See Exs. 18-19.) which shape is more often used for floor joists? Why? (Favor, Jonalyn) Formulas:

strength, ( 𝑠 = 𝑏𝑑 2 ) Stiffness, (𝑘 = 𝑏𝑑 3 )

Step 1: For 2 inches by 8 inches, solve for the strength and stiffness such that the breadth is 2 inches. Strength

Stiffness

𝑠 = 𝑏𝑑2 𝑠 = 8𝑖𝑛((2𝑖𝑛)2 ) 𝑠 = 32𝑖𝑛3 𝑘 = 𝑏𝑑3 𝑘 = 8𝑖𝑛((2𝑖𝑛)3 ) 𝑘 = 64𝑖𝑛4

d = 2 in b = 8 in

Step 2: For 2 inches by 8 inches, solve for the strength and stiffness such that the breadth is 8 inches. Strength

Stiffness

𝑠 = 𝑏𝑑2 𝑠 = 2𝑖𝑛((8𝑖𝑛)2 ) 𝑠 = 128𝑖𝑛3 d = 8 in

𝑘 = 𝑏𝑑3 𝑘 = 2𝑖𝑛((8𝑖𝑛)3 ) 𝑘 = 10244 b = 2 in b = 2 in

Step 3: For 4 inches by 6 inches, solve for the strength and stiffness such that the breadth is 6 inches. Strength

Stiffness

𝑠 = 𝑏𝑑2 𝑠 = 6𝑖𝑛((4𝑖𝑛)2 ) 𝑠 = 96𝑖𝑛3 𝑘 = 𝑏𝑑3 𝑘 = 6𝑖𝑛((4𝑖𝑛)3 ) 𝑘 = 384𝑖𝑛4

d = 4 in

b = 6 in

Step 4: For 4 inches by 6 inches, solve for the strength and stiffness such that the breadth is 4 inches. Strength

𝑠 = 𝑏𝑑2 𝑠 = 4𝑖𝑛((6𝑖𝑛)2 ) 𝑠 = 144𝑖𝑛3

Stiffness

𝑘 = 𝑏𝑑3 𝑘 = 4𝑖𝑛((6𝑖𝑛)3 ) 𝑘 = 864𝑖𝑛4

Step 5: Analyze the result

d = 6 in

b = 4 in

2 inches by 8 inches Breadth is 2 inches. Breadth is 8 inches. 4 inches by 6 inches Breadth is 6 inches. Breadth is 4 inches.

Strength

Stiffness

𝑠 = 32𝑖𝑛3 𝑠 = 128𝑖𝑛3

𝑘 = 64𝑖𝑛4 𝑘 = 1024𝑖𝑛4

𝑠 = 96𝑖𝑛3 𝑠 = 144𝑖𝑛3

𝑘 = 384𝑖𝑛4 𝑘 = 864𝑖𝑛4

21. Find the rectangle of maximum perimeter inscribed in a given circle. (Estares, Dave)

DD y x

We can see that there is a right triangle, so we can use the Pythagorean theorem: Diameter D is constant (circle is given): 𝑥 2 +𝑦 2 =𝐷2 Get the first derivative: 2x+2yy′=0 y′=−x/y Relate the first derivative above to the formula of the perimeter of the rectangle: Perimeter: P=2x+2y Get the first derivative: dP/dx=2+2y′=0 Substitute the value of y′: 2+2(−x/y)=0 y=x The largest rectangle is a square.

23. Find the most economical proportions for a covered box of fixed volume whose base is a rectangle with one side three times as long as the other. (Escanilla, Irish) Solution:

Use the formula in finding the volume for rectangle V = lwh V = (3x)(x)(y) V = 3x 2 y dV =0 dx

0 = 3𝑥 2 𝑦 0 = 6𝑥𝑦 + 3𝑥 2 𝑦′ 2𝑦 y’ = 𝑥 Total Area 𝐴 = 2𝑙𝑤 + 2𝑙ℎ + 2ℎ𝑤 = 2(3𝑥 2 ) + 2(3𝑥𝑦) + 2(𝑥𝑦) = 6𝑥 2 + 6𝑥𝑦 + 2𝑥𝑦 = 6𝑥 2 + 8𝑥𝑦 𝑑𝐴 𝑑𝑥

=0 0 = 12𝑥 + 8𝑦 + 8𝑥𝑦 ′ 0 = 12𝑥 + 8(𝑦 + 𝑥𝑦′)

*Substitute the value of y’ 0 = 12𝑥 + 8 [𝑦 + (𝑥) (−

2𝑦 𝑥

)]

2𝑦𝑥 0 = 12𝑥 + 8 [𝑦 − ( )] 𝑥 0 = 12𝑥 + 8(𝑦 − 2𝑦) 0 = 12𝑥 − 8𝑦 12𝑥 8

=

8𝑦

𝑦=

8 3𝑥 2

ANSWER :

𝟑 𝟐

times the shorter side of the base (x)

24. Solve Ex. 23 If the box has an open top. (Fortuna, Jarlhen) Let L=length (3x) W= width(x) H=height(y) Gven Volume: V= LxWxH V= (3x)(x)(y) V=3𝑋 2 y 0=6xy+3𝑥 2 y 3𝑥 2𝑦 −6𝑥𝑦 3𝑥 2

y’=

= 3𝑥 2

−2𝑦 𝑥

AREA: LW+2LH+2WH =3𝑥 2 +2(3xy)+2(xy) = 3𝑥 2 +6xy+2xy =3𝑥 2 +8xy 𝑑𝐴 𝑑𝑥

= 6x+8(xy’+y)=0 −2𝑦

=6x+8[x(-

𝑥

+y])=0

=6x+8(-2y+y)=0 =6x+8(-y)=0 =6x-8y=0 6𝑥 8𝑦 8 𝟑

=8

=y

𝟒

25. Find the most economical proportions of a quart can. (Buenconsejo, Dominic)

Volume: 1

𝑉 = 4 𝜋𝑑2 ℎ = 1 𝑞𝑢𝑎𝑟𝑡

Differentiating the equation using Product Rule, we would have: 1 4

𝑑ℎ

𝜋 [𝑑2 𝑑𝑑 + 2𝑑ℎ] = 0

𝜋𝑑2 𝑑ℎ −2𝜋𝑑ℎ ( ) = 4 𝑑𝑑 4 𝑑ℎ −2ℎ = 𝑑𝑑 𝑑

Total Area: 𝜋𝑑 2 𝐴𝑇 = 2 ( ) + 𝜋𝑑ℎ 4 Simplifying the equation: 𝜋𝑑2 𝐴𝑇 = + 𝜋𝑑ℎ 2 Getting the derivative of the total area with respect to the diameter, we will use Power Rule and Product Rule. 𝑑𝐴𝑇 𝑑ℎ = 𝜋𝑑 + 𝜋 [𝑑 + ℎ] = 0 𝑑𝑑 𝑑𝑑 Simplifying the equation: 𝑑ℎ 𝜋𝑑 + (𝜋𝑑) ( ) + 𝜋ℎ = 0 𝑑𝑑 Dividing both sides by 𝜋: 𝑑ℎ

𝑑 + 𝑑 (𝑑𝑑) + ℎ=0 Substituting the value of

𝑑ℎ 𝑑𝑑

which is

−2ℎ 𝑑

to the equation:

−2ℎ 𝑑+𝑑( )+ℎ = 0 𝑑 𝑑−ℎ =0 𝑑=ℎ The most economical proportions of a quart can is diameter 𝒅 = height 𝒉.

26. Find the most economical proportions for a cylindrical cup. (Duazo, Rigen) Step 1: Make an illustration.

Step 2: Write the formula for its volume. 1

𝑉 = 4 𝜋𝑑2 ℎ Step 3: Find the derivative of ℎ with respect to 𝑑. (Use the product rule) 𝑑ℎ

Since that 𝑑𝑑 = 0, then we have:

1

𝑑ℎ

0 = 4 𝜋 (𝑑2 𝑑𝑑 + 2𝑑ℎ) 1

𝑑ℎ

1

1

𝑑ℎ

1

1

𝑑ℎ

1

0 = 4 𝜋𝑑2 𝑑𝑑 + 4 𝜋2𝑑ℎ 0 = 4 𝜋𝑑2 𝑑𝑑 + 4 𝜋2𝑑ℎ 0 = 4 𝜋𝑑2 𝑑𝑑 + 4 𝜋2𝑑ℎ 1 4

𝜋𝑑2

𝑑ℎ 𝑑𝑑

1

= 𝜋2𝑑ℎ 4

1

Divide both sides with 4 𝜋𝑑2 . We have: 1 𝑑ℎ 1 𝜋𝑑2 = − 𝜋2𝑑ℎ 4 𝑑𝑑 4 1 𝜋𝑑2 4

𝑑ℎ 𝑑𝑑

=−

2ℎ 𝑑

Step 4: Use its area considering that one end is open to find the derivative of 𝐴 with respect to 𝑑. Use product rule. Formula: 1 𝐴𝑡𝑜𝑝 𝑐𝑜𝑣𝑒𝑟+𝑐𝑦𝑙𝑖𝑛𝑑𝑟𝑖𝑐𝑎𝑙 𝑐𝑢𝑝 𝑤𝑖𝑡ℎ𝑜𝑢𝑡 𝑡𝑜𝑝 𝑎𝑛𝑑 𝑏𝑜𝑡𝑡𝑜𝑚 𝑐𝑜𝑣𝑒𝑟 = 𝜋𝑑 2 + 𝜋𝑑ℎ 4 𝑑𝐴 𝑑𝑑 1 2

1

𝑑ℎ

= 2 𝜋𝑑 + 𝜋 (𝑑 𝑑𝑑 + ℎ) = 0

𝜋𝑑 + 𝜋𝑑 𝑑ℎ

𝑑ℎ 𝑑𝑑 1

+ 𝜋ℎ = 0

𝜋𝑑 𝑑𝑑 = − 2 𝜋𝑑 − 𝜋ℎ

Divide both sides with 𝜋𝑑. 𝜋𝑑

𝑑ℎ 1 =− 𝜋𝑑−𝜋ℎ 𝑑𝑑 2

𝜋𝑑 𝑑ℎ 𝑑𝑑

1

= 2𝑑 − ℎ 1

ℎ = 2 𝑑 or ℎ = 𝑟 ANSWER: Height should be equal to the radius of its base.

27. Find the most economical proportions for a box with an open top and a square base. (Atutubo, Jhyne) Volume: V=x2y 0 = x2y′+2xy y′ = −2y/x Area: A=x2+4xy dA/dx=2x+4(xy′+y)=0 2x+4[x(−2y/x)+y]=0 2x−8y+4y=0 2x=4y x=2y base = 2 × altitude

answer

28. The perimeter of an isosceles triangle is P inches. Find the maximum area. (Enaje, Christian) Perimeter: P = 2a + b a=y,b=x P = 2y + x Equate to zero and derive: 0 = 2y′+1 −1 y′ = 2 = .5

y

y h

x

From the figure: ℎ = √𝑦 2 − .25𝑥 2 Area: 1 A=2 𝑥ℎ 1 𝐴 = 𝑥√𝑦 2 = .25𝑥 2 2 𝑑𝐴 𝑑𝑥

[

2(𝑥𝑦𝑦 ′ −.25𝑥 2 ) 2√𝑦 2 −.25𝑥 2

=

1

[𝑥 ( 2

2𝑦𝑦 ′ −.5𝑥

2√𝑦 2 −.25𝑥 2

+ √𝑦 2 − .25𝑥 2 )] = 0

+ √𝑦 2 − .25𝑥 2 = 0] (√𝑦 2 − .25𝑥 2 )

𝑥𝑦𝑦 ′ − .25𝑥 2 + 𝑦 2 − .25𝑥 2 = 0 𝑥𝑦(−0.5) − 0.5𝑥2 + 𝑦2 = 0 𝑦2 − 0.5𝑥𝑦 − 0.5𝑥2 = 0 2𝑦2 − 𝑥𝑦 − 𝑥2 = 0 Solving for y by quadratic formula: a = 2; b = -x; c = −𝑥 2 𝑦= 𝑦=

−𝑏 ± √𝑏 2 − 4𝑎𝑐 2𝑎

−(−𝑥) + √𝑥 2 − 4(2)(−𝑥 2 ) 2(2) 𝑥 ± √𝑥 2 + 8𝑥 2 𝑦= 4 𝑦=

𝑥 ± 3𝑥 4

1 𝑦 = 𝑥 𝑎𝑛𝑑 − 𝑥 2 Use y = x Therefore: P = 2x + x = 3x 1

x = y = 3𝑃 ℎ = √𝑦 2 − .25𝑥 2 1 1 1 ℎ = √( 𝑃)2 − ( 𝑃)2 3 4 3 1 1 ℎ = √ 𝑃2 − 𝑃2 9 36 1 ℎ = √ 𝑃2 12

ℎ= √

1 2√3

𝑃2

1 𝑥ℎ 2 1 1 1 𝐴𝑚𝑎𝑥 = ( 𝑃)( 𝑃) 2 3 2√3 1 𝐴𝑚𝑎𝑥 = 𝑃2 12√3 𝐴=

𝐴𝑚𝑎𝑥 = 𝐴𝑚𝑎𝑥 =

1 12√3

∗

√3 √3

𝑃2

√3 2 𝑃 36

30. Find the proportion of the circular cylinder of the largest volume that can be inscribed in a given sphere. (Escanilla, Irish)

Solution: 𝐷 2 = 𝑑 2 + ℎ2 0 = 2𝑑 + 2ℎ

𝑑ℎ 𝑑𝑑

𝑑ℎ 𝑑 = − 𝑑𝑑 ℎ Find the volume of the cylinder: 1 2 𝜋𝑑 ℎ 4 𝑑𝑉 𝜋 2 𝑑ℎ = [𝑑 + 2𝑑ℎ] = 0 𝑑ℎ 4 𝑑𝑑 𝑉=

𝑑

𝑑ℎ + 2ℎ = 0 𝑑𝑑 𝑑ℎ

*substitute the value of 𝑑𝑑 to the equation 𝑑 𝑑 (− ) + 2ℎ = 0 ℎ −𝑑 2 = −2ℎ ℎ (−2ℎ2 = −𝑑 2 )(−1) 𝑑 2 = 2ℎ2

√𝑑2 = √2ℎ2 𝑑 = √2ℎ ANSWER: √2 times the height (h)

31. In Ex. 30 find the shape of the cylinder if its convex surface area is to be a maximum. (Fortuna, Jarlhen) Let d= diameter and h= height A=dh 𝑑𝐴

𝑑ℎ

=𝑑𝑑(d𝑑𝑑+h)=0 𝑑ℎ

=d𝑑𝑑+h=0 From solution 30 𝑑ℎ

=d-(𝑑𝑑 + ℎ)=0 h=𝑑2 h √ℎ2 =√𝑑 2 h=d diameter = height 32. Find the dimension of the largest rectangular building that can be placed on a right-triangular lot, facing one of the perpendicular sides. (Buenconsejo, Dominic)

Area of rectangle: 𝐴 = 𝑥𝑦

Using ratio and proportion, based on the figure: 𝑦 𝑏 = 𝑎−𝑥 𝑎 𝑦=

𝑏 (𝑎 − 𝑥) 𝑎

From the formula for finding the area: 𝐴 = 𝑥𝑦 𝑏 𝐴 = 𝑥 (𝑎 − 𝑥) 𝑎 Simplifying the equation: 𝑏 𝐴 = 𝑏𝑥 − 𝑥 2 𝑎

Getting the derivative of the area with respect to the x, we will use Power Rule. 𝑑𝐴 2𝑏 =𝑏− 𝑥=0 𝑑𝑥 𝑎 2𝑏 𝑥=𝑏 𝑎 𝑎 𝑥= 2 𝑏

Since 𝑦 = 𝑎 (𝑎 − 𝑥), then 𝑦=

𝑏 𝑎 (𝑎 − ) 𝑎 2

𝑦=𝑏− 𝑦=

𝑏 2

𝑏 2 𝒂

𝒃

The dimensions of the largest rectangular building are 𝟐 by 𝟐 units.

33. A lot has the form of a right triangle, with perpendicular sides 60 and 80 feet long. Find the length and width of the largest Rectangular building that can be erected, facing the Hypotenuse of the triangle. (Duazo, Rigen) Step 1: Make an illustration.

Step 2: Find the area of the rectangular building whose width is x, and length is y. 𝐴 = 𝑥𝑦 Considering that by similar triangle, we have: cot 𝛼 =

𝑟 𝑥

60

= 80

𝑟 𝑥

60 80

=

20

𝑟

3

𝑥 (𝑥) = (4) 𝑥 𝒓= 𝑠

𝟑 𝒙 𝟒

80

cot 𝛽 = 𝑥 = 60 𝑠 𝑥

=

𝑠

80 60

20

4

𝑥 (𝑥) = (3) 𝑥 𝟒

𝒔 = 𝟑𝒙 Now, we found the value for 𝑟 and 𝑠. Since that we have the value for the hypotenuse of the right triangle by using the Pythagorean theorem: √𝑎2 + 𝑏 2 = 𝑐 2 √602 + 802 = 10,000 𝑐 2 =10,000 𝑐 = √10000 𝒄 = 𝟏𝟎𝟎 We proceed with: Step 3: Determine the value of 𝑦 using the solved value for 𝑟 and 𝑠. We have, 𝑟 + 𝑦 + 𝑠 = 100 •

3

4

Substitute 4 𝑥 to 𝑟 and 3 𝑥 to 𝑠. 3 4 𝑥 + 𝑦 + 𝑥 = 100 4 3 100 = 𝑦 +

•

9+16 12

25

𝑥 = 𝑦 + 12 𝑥

Isolate the 𝑦 variable, which is the length of the rectangular building. 𝑦 = 100 −

After solving for 𝑦, we now proceed with: Step 4: Find the derivative of 𝐴 with respect to 𝑥. Using the formula for the Area of Rectangular building 𝐴 = 𝑥𝑦 25

and substituting the value 100 − 12 𝑥 for 𝑦, we have:

25 𝑥 12

25

𝐴 = 𝑥 (100 − 12 𝑥) 25

𝐴 = 100𝑥 − 12 𝑥 2 𝑑𝐴 𝑑𝑥 𝑑𝐴 𝑑𝑥

25

= 100(1) − 2(12)𝑥 2−1 50

= 100 − 12 𝑥 •

Isolate the value of 𝑥. 50 12

𝑥 = 100

50 𝑥=100 12 50 12

𝑥= 𝑥= 𝑥=

100 50 12

100 (12) 50 1,200 50

𝒙 = 𝟐𝟒 𝒇𝒕. Step 5: Substitute 𝑥 = 24 𝑓𝑡. to solve the value for 𝑦. (Refer to step 3) 25

𝑦 = 100 − 12 𝑥 25

𝑦 = 100 − 12 (24 𝑓𝑡. ) 𝑦 = 100 −

600 𝑓𝑡. 12

𝑦 = 100 − 50 𝑓𝑡. 𝒚 = 𝟓𝟎 𝒇𝒕.

Answer: Dimension is 50 ft. by 24 ft.

34. Solve Problem 34 above if the lengths of the perpendicular sides are a, b. (Atutubo, Jhyne) Area: A= xy

By similar triangle: cotα=r/x=b/a r = bx/a cotβ = s/x =a/b s = ax/b r + y + s = √𝑎2 + 𝑏 2 ; Substitute the values of r and s in this formula. bx/a + y + ax/b = √𝑎2 + 𝑏 2 y+

𝑥(𝑎2 + 𝑏2 ) 𝑎𝑏

= √𝑎2 + 𝑏 2

y = √𝑎2 + 𝑏 2 −

𝑥(𝑎2 + 𝑏 2 ) 𝑎𝑏

y = √𝑎2 + 𝑏 2 (1 −

𝑥√𝑎2 + 𝑏2 𝑎𝑏

A = x (√𝑎2 + 𝑏 2 (1 − 𝑥√𝑎2 + 𝑏2

A=

𝑎𝑏

dA/dx =

); substitute this value to the formula A = xy. So,

𝑥√𝑎2 + 𝑏 2 𝑎𝑏

))

(𝑎𝑏 −x√𝑎2 + 𝑏 2 ) =

√𝑎2 + 𝑏2 𝑎𝑏

√𝑎2 + 𝑏2 𝑎𝑏

(𝑥𝑎𝑏 − 𝑥 2 √𝑎2 + 𝑏 2 )

(𝑎𝑏 − 2𝑥√𝑎2 + 𝑏 2 ) = 0

𝑎𝑏 − 2𝑥√𝑎2 + 𝑏 2 = 0 𝑎𝑏 2√𝑎2 + 𝑏 2

=𝑥

For y: y = √𝑎2 + 𝑏 2 (1 − y=

√𝑎2 + 𝑏2

𝑦=

𝑎𝑏

(𝑎𝑏 −

𝑥√𝑎2 + 𝑏 2 𝑎𝑏 𝑎𝑏

2√𝑎2 + 𝑏2

)=

√𝑎2 + 𝑏2 𝑎𝑏

√𝑎2 + 𝑏 2 ) =

(𝑎𝑏 − 𝑥√𝑎2 + 𝑏 2 );

√𝑎2 + 𝑏2 𝑎𝑏

(𝑎𝑏 −

𝑎𝑏

)= 2

𝑎𝑏 2√𝑎2 + 𝑏2

√𝑎2 + 𝑏2 𝑎𝑏

=𝑥 𝑎𝑏

(2)

1 √𝑎2 + 𝑏 2 2

So, the dimension of the rectangle is A = xy. A=

𝑎𝑏 2√𝑎2 + 𝑏2

𝑥

1 2

√𝑎2 + 𝑏 2

35. A page is to contain 24 sq. in. of print. The margins at top and bottom are 1.5 inch, at the sides 1 inch. Find the most economical dimensions for the page. (Enaje, Christian) W= x-2, L= y-3,

A = LW=24 To get y: 24 = (y-3) (x-2) 24

a.)(𝑥−2) =

(𝑦−3)(𝑥−2) (𝑥−2)

24

b.) (𝑥−2) = 𝑦 − 3 24

c.) 3 + (𝑥−2) = 𝑦

The Area of the Page A= x*y 24 A=x(3 + 𝑥−2) 24𝑥

A=3𝑥 + 𝑥−2 A’= 3 +

(24)(𝑥−2)−(24𝑥)(1)

A’= 3 +

(𝑥−2)2 (24𝑥−48−24𝑥) (𝑥−2)2 48

A’= 3 - (𝑥−2)2 48

0 = 3 – (𝑥−2)2 48

3 = (𝑥−2)2 3 (𝑥 − 2)2 = 48 3(𝑥 − 2)2 48 = 3 3 √(𝑥 − 2)2 = √16 x-2 = 4 x=6 24

B.) y = 3 + 𝑥−2 y=3+

24

6−2

y=3+6 y=9 36. A Norman window consists of a rectangle surmounted by a semicircle. What shape gives the most light for the given perimeter? (Enaje, Christian) 𝐺𝑖𝑣𝑒𝑛 𝑝𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟: 𝑃 = 𝑏 + 2(ℎ − 𝑟) + 𝜋𝑟 𝑃 = 𝑏 + 2ℎ − 2𝑟 + 𝜋𝑟 𝑊ℎ𝑒𝑟𝑒: 𝑏 = 2𝑟 𝑟 = 12𝑏 Thus, 1 𝑃 = 𝑏 + 2ℎ − 𝑏 + 𝜋𝑏 2 1 𝑃 = 2ℎ + 𝜋𝑏 2 𝑑𝑃 𝑑ℎ 1 =2 + 𝜋=0 𝑑𝑏 𝑑𝑏 2 𝑑ℎ 1 =− 𝜋 𝑑𝑏 4 Light is most if area is maximum: 1 𝐴 = 𝜋𝑟 2 + 𝑏(ℎ − 𝑟) 2 1 1 1 𝐴 = 𝜋( 𝑏)2 + 𝑏(ℎ − 𝑏) 2 2 2

1 2 1 𝜋𝑏 + 𝑏ℎ − 𝑏 2 8 2 1 𝐴 = (𝜋 − 4)𝑏 2 + 𝑏ℎ 8 𝑑𝐴 2 𝑑ℎ = (𝜋 − 4)𝑏 + 𝑏 +ℎ =0 𝑑𝑏 8 𝑑𝑏 1 1 𝜋𝑏 − 𝑏 − 𝜋𝑏 + ℎ = 0 4 4 ℎ=𝑏 Breadth = Height 𝐴=

37. Solve Problem 36 above if the semicircle is stained glass admitting only half the normal amount of light. (Atutubo, Jhyne)

Given perimeter: P=b+2(h−r)+πr P=b+2h−2r+πr Where: b=2r r= b/2 Thus, P=b+2h−b+1/2(πb) P=2h+1/2(πb) dP/db = 2 dh/db + ½( π) = 0 dh/db = -1/4 π Half amount of light is equivalent to half of the area of the semicircle. A = ¼ π𝑟 2 + b(h − r) 𝑏 2

b

= ¼ π(2) + b (h − 2) = ¼ π (𝑏 2 /4)+ bh – 2

= 1/16 π𝑏 + bh – A=

𝑏2 16

𝑏2

𝑏2

2

2

(π − 8) + 𝑏ℎ 2𝑏

dA/db = 16 (π − 8) +b(dh/db)+h = 0; Substitute the value of dh/db in this equation. 1/8(πb)−b−1/4πb+h=0 -1/8 πb -b = -h h= b(1/8 π +1) So height is (1/8 π +1) of the breadth.