Welcome to our community

Be a part of something great, join today!

torres' question at Yahoo! Answers regarding minimizing the cost of a box

  • Thread starter
  • Admin
  • #1

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Here is the question:

CALCULUS question HELP?

I want to make a large box with a square base that has a volume of 6000in^3. The material for the top and bottom costs 5 cents per square inch and the side material costs 3 cents per sq in. Make a function C that represents the cost of the box in terms of the side length of the bottom of the box. Find the one critical number for C and then give the dimensions of the cheapest box.

I have a final for calculus so I would like to know the steps on how to find the answer. If you could do that it would really help. Thank you in advance.
I have posted a link there to this thread so the OP can view my work.

edit: Before I could post my solution the OP deleted the question.
 
  • Thread starter
  • Admin
  • #2

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Hello torres,

If we let $x$ be the side length of the bottom of the box and $h$ be the height of the box, then the constant volume V of the box is:

\(\displaystyle V=hx^2\)

This is our constraint.

If we then let $c_1$ be the cost per unit area of the top/bottom and $c_2$ be the cost per unit area of the sides, then our total cost function $C$ is:

\(\displaystyle C(h,x)=2c_1x^2+4c_2hx\)

This is our objective function.

Now, if we solve the constraint for $h$, we obtain:

\(\displaystyle h=\frac{V}{x^2}\)

Substituting for $h$ into the objective function, we obtain:

\(\displaystyle C(x)=2c_1x^2+4c_2Vx^{-1}\)

Now, to find the critical value(s), we should differentiate with respect to $x$ and equate the result to zero:

\(\displaystyle C'(x)=4c_1x-4c_2Vx^{-2}=\frac{4\left(c_1x^3-c_2V \right)}{x^2}=0\)

This implies:

\(\displaystyle c_1x^3-c_2V=0\implies x=\left(\frac{c_2}{c_1}V \right)^{\frac{1}{3}}\)

We can see that for values of $x$ less than this critical value the first derivative is negative and for values larger than this critical value the first derivative is positive, therefore by the first derivative test we may conclude that this critical value is at a relative minimum.

Thus the height of the box at this critical value is:

\(\displaystyle h=\frac{V}{\left(\frac{c_2}{c_1}V \right)^{\frac{2}{3}}}=\left(\left(\frac{c_1}{c_2} \right)^2V \right)^{\frac{1}{3}}=\frac{c_1}{c_2}x\)

Now, plugging in the given data:

\(\displaystyle V=6000\text{ in}^3,\,c_1=5\frac{\text{¢}}{\text{in}^2},\, c_2=3\frac{\text{¢}}{\text{in}^2}\)

We find:

\(\displaystyle x=\left(\frac{3}{5}\cdot6000 \right)^{\frac{1}{3}}\text{ in}=2\sqrt[3]{450}\text{ in}\)

\(\displaystyle h=\frac{5}{3}\cdot2\sqrt[3]{30}\text{ in}=\frac{10}{3}\sqrt[3]{450}\text{ in}\)