1. ## Designing a Tank

Your iron works has contracted to design and build a $500ft^3$, square based, open top, rectangular steel holding tank for a paper company. The tank is to be made by welding thin stainless steel plates together along their edges. As the production engineer, your job is to find dimensions for the base and height that will make the tank weigh as little as possible.

These optimiziation problems are really getting to me, I have no idea where to start. Could anyone help solve it and/or provide tips on how I could get started. Thanks for the help.

2. Let $x = length, y = width, z = height$.

The Volume, $V = xyz = 500$.

The Surface Area, $A = xy + 2zy + 2xz$. (Notice it's not $2xy$ because there is no top.)

We want surface area to be minimized, since that will minimize weight.

Can you take it from here with this system of equations?

Hint: What does "square based" tell you about $x$ and $y$?

3. X and Y will be equal, will they not?

4. Yes.

So we preserve $x$.

$V = x^2z = 500$

$A = x^2 + 2zx + 2zx = x^2 + 4zx$

We will obviously use A to minimize weight, but it has a $z$ in it. So, get rid of the $z$.

5. Do derivatives come into play now?

6. Of course.

We first want to eliminate $z$, though (this is one variable calculus after all).

So, we rearrange and get $z = \frac{500}{x^2}$

Substituting this into the Area equation, we get

$A = x^2 + 4zx = x^2 + \frac{2000}{x}$

Using the first derivative test, we see the minimum of this is at $x = 10$.

Since, $x = y$, we have $y = 10$.

Solving for $z$, we get $z = 5$.

So we have our dimensions. Get it?

7. I took the first derivative of the SA and came up with $2x-2000/x^2$. How could x be 10 in this case? Unless I did it wrong...

We wish to find the critical points of $A = x^2 + \frac{2000}{x^2}$. Using your correct derivative,

$A' = 2x - \frac{2000}{x^2}$. We set this equal to 0 and solve for $x$.

$A' = 2x - \frac{2000}{x^2} = 0$.

$2x = \frac{2000}{x^2}$

$x = \frac{1000}{x^2}$

$x^3 = 1000$

$x = 10$

$A'$ is negative before $x = 10$ and $A'$ is positive after $x = 10$.

So $x = 10$ is our minimum.

9. Ok, it makes sense now, thanks for the help.