# Torres' question at Yahoo Answers regarding minimizing the cost of a box

• MHB
• MarkFL
In summary: Therefore the cheapest box is a cube with side length $2\sqrt[3]{450}\text{ in}$ and height $\frac{10}{3}\sqrt[3]{450}\text{ in}$. In summary, we can find the cheapest box with a volume of 6000in^3 by setting up a cost function and differentiating it to find the critical value. Plugging in the given data, we can determine the dimensions of the cheapest box to be a cube with side length $2\sqrt[3]{450}\text{ in}$ and height $\frac{10}{3}\sqrt[3]{450}\text{ in}$.
MarkFL
Gold Member
MHB
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.

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:

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

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

## What is Torres' question about minimizing the cost of a box?

Torres' question at Yahoo Answers is about finding the minimum cost of a box given that the material for the base costs $3 per square foot and the material for the sides costs$2 per square foot.

## How can I calculate the minimum cost of a box?

To calculate the minimum cost of a box, you need to find the dimensions of the box that will minimize the surface area. This can be done by finding the derivative of the surface area function and setting it equal to 0. Then, solve for the dimensions that give a minimum value.

## What is the formula for the surface area of a box?

The formula for the surface area of a box is SA = 2lw + 2lh + 2wh, where l, w, and h are the length, width, and height of the box, respectively.

## How does the cost of the materials affect the minimum cost of the box?

The cost of the materials directly affects the minimum cost of the box. In this case, the cost of the base material is twice the cost of the side material, so the minimum cost will occur when the base is twice the size of the sides.

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