Question

A closed box with a square base is to have a volume of 171 comma 500 cm cubed. The material for the top and bottom of the box costs ​$10.00 per square​ centimeter, while the material for the sides costs ​$2.50 per square centimeter. Find the dimensions of the box that will lead to the minimum total cost. What is the minimum total​ cost? Write an equation for​ C(x), the cost of the box as a function of​ x, the length of a side of the base. ​C(x)equals nothing

Answer 1

Answer:

$C(x)=\dfrac{20x^3+1715000}{x}\\ Minimum cost, C(35)=\ 29,400$

The dimensions that will lead to minimum cost of the box are a base length of 35 cm and a height of 140 cm.

Step-by-step explanation:

Volume of the Square-Based box=171,500 cubic cm

Let the length of a side of the base=x cm

Volume $=x^2h$

$x^2h=171,500\\h=\dfrac{171500}{x^2}$

The material for the top and bottom of the box costs ​$10.00 per square​ centimeter.

Surface Area of the Top and Bottom $=2x^2$

Therefore, Cost  of the Top and Bottom $=\ 10X2x^2=20x^2$

The material for the sides costs ​$2.50 per square centimeter.

Surface Area of the Sides=4xh

Cost of the sides=$2.50 X 4xh =10xh

$\text{Substitute h} =\dfrac{171500}{x^2} into 10xh\\Cost of the sides=10x(\dfrac{171500}{x^2})=\dfrac{1715000}{x}$

Therefore, total Cost of the box

$= 20x^2+\dfrac{1715000}{x}\\C(x)=\dfrac{20x^3+1715000}{x}$

To find the minimum total cost, we solve for the critical points of C(x). This is obtained by equating its derivative to zero and solving for x.

$C'(x)=\dfrac{40x^3-1715000}{x^2}\\\dfrac{40x^3-1715000}{x^2}=0\\40x^3-1715000=0\\40x^3=1715000\\x^3=1715000\div 40\\x^3=42875\\x=\sqrt[3]{42875}=35$

Recall that:

$h=\dfrac{171500}{x^2}\\Therefore:\\h=\dfrac{171500}{35^2}=140cm$

The dimensions that will lead to minimum costs are base length of 35cm and height of 140cm.

Therefore, the minimum total cost, at x=35cm

$C(35)=\dfrac{20(35)^3+1715000}{35}=\ 29,400$