Question

find the dimensions of the crate that will minimize the total cost of material

Answer 1

Let's say that the rectangular shipping crate with a square base has the dimensions:

$a\times a\times b$

The volume of such a shipping crate is:

$V=a\cdot a\cdot b=a^2\cdot b$

Additionally, the lateral sides have an area of:

$A_L=a\cdot b$

For the base and the top:

$A_B=A_T=a^2$

The cost of the material is $7 per square foot for the base and $2 per square foot for the top and the sides. Then, the total cost C will be:

$\begin{gathered} C=2\cdot4\cdot A_L+7\cdot A_B+2\cdot A_T=8A_L+9A_B \\ C=8a\cdot b+9a^2 \end{gathered}$

Because there are 4 sides, one base, and one top, and the areas of the base and the top are equal. We know that the volume is a fixed value, then:

$a^2\cdot b=1152\Rightarrow b=\frac{1152}{a^2}$

Using this result in the expression of the cost:

$C=8\cdot a\cdot\frac{1152}{a^2}+9a^2=\frac{9216}{a}+9a^2$

Now, to find the minimum cost, we take the derivative of C with respect to a:

$\frac{dC}{da}=-\frac{9216}{a^2}+18a$

The minimum is obtained by solving the equation dC/da = 0:

$\begin{gathered} -\frac{9216}{a^2}+18a=0 \\ 18a=\frac{9216}{a^2} \\ a^3=\frac{9216}{28} \\ a^3=512 \\ a=8 \end{gathered}$

Using this result, we can calculate b:

$b=\frac{1152}{a^2}=\frac{1152}{8^2}=\frac{1152}{64}=18$

Finally, the dimensions of the crate that will minimize the total cost of material are:

$8ft\times8ft\times18ft$