Question

A rectangular box is to have a square base and a volume of 72 ft3. If the material for the base costs $0.62/ft2, the material for the sides costs $0.15/ft2, and the material for the top costs $0.18/ft2, determine the dimensions (in ft) of the box that can be constructed at minimum cost.
a. length
b. width
c. height

Answer 1

Answer:

a. length  = 0.7211 ft

b. width  = 0.7211 ft

c. height = 140.3846 ft

Step-by-step explanation:

This is an optimiztion with restriction problem.

We have to minimize the cost, with the restriction of the volume being equal to 72 ft3.

As the cost for the sides is constant, we know that length and width are equal.

Then, we can express the volume as:

$V=x\cdot y\cdot z=x^2z=73$

being x: length and z: height

We can express the height in function of the length as:

$x^2z=73\\\\z=73x^{-2}$

Then, the cost of the box can be expressed as:

$C=0.62(x^2)+4*0.15(xz)+0.18(x^2)=(0.62+0.18)x^2+0.60xz\\\\C=0.8x^2+0.60x*x^{-2}=0.8x^2+0.6x^{-1}$

To optimize C, we derive and equal to zero

$\dfrac{dC}{dx}=\dfrac{d}{dx}[0.8x^2+0.6x^{-1}]=1.6x-0.6x^{-2}=0\\\\\\1.6x=0.6x^{-2}\\\\x^{1+2}=0.6/1.6=0.375\\\\x=\sqrt[3]{0.375} =0.7211$

The height z is then

$z=73x^{-2}=\dfrac{73}{0.7211^2}=\dfrac{73}{0.52}=140.3846$