Question

A rectangular box with a volume of 784ft^3 is to be constructed with a square base and top. The cost per square foot for the bottom is 20 cents, for the top is 16 cents, and for the sides is 1.5 cents. What dimensions will minimize the cost? The length of one side of the base is ? The height of the box is?

Answer 1

Let's dimension the horizontal length with the name X and the vertical dimensions as Y.

In this way the total volume will be given under the function

$V = x^2 y$

The cost for the bottom is given by $(x^2)(20)=20x^2$

While the cost for performing the top by $(x^2)(16) = 16x^2$

The cost for performing the sides would be given by $(1.5)(xy)(4) = 6xy$

Therefore the total cost would be

$c_{total}= 36x^2 +6xy$

The total volume is equivalent to

$784 = x^2y$

$xy = \frac{784}{x}$

Replacing in our cost function

$c_{total}= 36x^2 +6xy$

$c_{total}= 36x^2 +6(\frac{784}{x})$

Obtaining the first derivative and equalizing to zero we will obtain the ideal measure, therefore

$c' = 0$

$c' = 72x-\frac{4707}{x^2}$

$0= 72x-\frac{4707}{x^2}$

$x = (\frac{523}{2})^{1/3}$

$x = 6.3947ft$

Then,

$784 = x^2y$

$y = \frac{784}{x^2}$

$y = \frac{784}{6.3947^2}$

$y = 19.17ft$

In this way the measures of the base should be 6.3947ft (width and length) and the height of 19.17ft.