A pencil cup with a capacity of 40 in.3 is to be constructed in the shape of a rectangular box with a square base and an open top.

Question

A pencil cup with a capacity of 40 in.³ is to be constructed in the shape of a rectangular box with a square base and an open top. If the material for the sides costs 17¢/in.² and the material for the base costs 41¢/in.², what should the dimensions of the cup be to minimize the construction cost (in ¢)?

Let x be the length (in in.) of the side of the square base of the open box, and h be the height (in in.) of the box.

x =  in. (Round your answer to two decimal places, if necessary.)

Complete the following parts.

(a)     Give a function f in the variable x for the quantity to be optimized.

f(x) = 

 

 cents

(b)     State the domain of this function. (Enter your answer using interval notation.)

 

(c)     Give the formula for h in terms of x.

Answer 1

Volume = V = 40 = h(x^2) = height times side of the bottom squared

h = 40/x^2

x^2 costs 41 cents per square inch

xh costs 17 cents per square inch

total surface area = x^2 +4xh = x^2 +4(40)/x^2

total cost =f(x) = 41x^2 + 17(4)(40)/x^2 = 41x^2 +2720/x^2

take the derivative and set =0

f'(x) = 82x - 8160/x^3 =0

multiply by x^3

82x^4 = 8160

x^4 = 8160/82

x = the 4th root of 8160/82

x= about 3.16 inches

Minimum Cost = .41[8160/82]^(1/2) + 4(.17)(40/(8160/82)^(1/2)

C= about .41(3.16)^2 +.17(160)/(3.16)^2

= 4.094 + 2.727

= about $6.82 per open top box

domain for x is 0<x<infinity (0, infinity)

as x approaches infinite inches, h approaches zero inches and cost approaches infinite dollars and cents

as x approaches zero inches, h appoaches infinite inches and cost approaches infinite dollars and cents