Question

A rectangular storage container with open top is to have a volume of 50 ft3 . The length of the base of the container is to be twice its width. Material for the sides costs $5 per square foot, and materials for the base cost $8 per square foot. Find the dimensions that minimize the cost of the material for the box.

Answer 1

Answer:

Length = 2.862 ft

Width = 1.431 ft

Height = 3.05 ft

Explanation:

Let L be lenght, W be width, and H be height of the storage container. Since length is to be twice the width, it means L = 2W. The volumn of the container would also be:

$V = LWH = L(2L)H = 2L^2H$

SInce the volumn constraint is 50, that means:

$2L^2H = 50$

$H = \frac{25}{L^2}$

The cost for the base would be its area times unit price

$C_b = A_bU_b = LWU_b = 2L^2U_b = 2*8L^2 = 16L^2$

Likewise, the cost for the sides would be

$C_s = A_sU_s = 2(L+W)HU_s = 6LH5 = 30LH$

We can substitute H from the equation above:

$C_s = 30L\frac{25}{L^2} = \frac{750}{L}$

Therefore the total material cost

$C = C_b + C_s = 16L^2 + \frac{750}{L}$

To find the minimum of this function, we can take first derivative then set to 0:

$C^{'} = (16L^2)^{'} + (\frac{750}{L})^{'} = 0$

$32L - \frac{750}{L^2} = 0$

$32L = \frac{750}{L^2}$

$L^3 = 750/32 = 23.4375$

$L = 23.4375^{1/3} \approx 2.862ft$

So W = L/2 = 1.43ft and

$H = \frac{25}{L^2} = 3.05 ft$

If we take the 2nd derivative and substitute L = 2.862 we would have

$C{''} = 32 + 2\frac{750}{L^3} = 32 + \frac{1500}{23.43} = 96 > 0$

Hence L = 2.862 would yield the minimum material cost