# Thread: More Optimization and Modeling

1. ## More Optimization and Modeling

A rectangular beam is cut from a cylindrical log of radius 30 cm. The strength of a beam of width w and height h is proportional to $wh^2$. Find the width and height of the beam of maximum strength.

I'm not sure what to do to set this up... After the setup, It'll be easy.. I'll just take the derivative of the strength equation to maximize it...

All i have so far is $s = kwh^2$. I don't know how to write any other equations to help with writing one variable in terms of another for substitution.

EDIT:
I had one of those light bulbs pop up in my head while i was standing my the microwave waiting for my hot pockets to heat up

Here's my work... please check to see if i thought of the correct thing.

since r = 30, then d = 60.
the diameter is also a diagonal of the rectangle so:
$60^2 = h^2 + w^2$

Solve for h.
$h = \sqrt{3600-w^2}$

Now, Since the strength function is $s = kwh^2$, I can substitute.
$s = kw(\sqrt{3600-w^2})^2$
$s = 3600kw - kw^3$

Now I will take the derivative and set it equal to 0 to find where the strength is maximized:
$s' = 3600k - 3kw^2 = 0$
$3k(1200-w^2)=0$
$w = \sqrt{1200}$

Now the 2nd derivative test to be certain it is a maximum.
$s'' = -6kw$
$s''(\sqrt{1200}) = -6k(\sqrt{1200})$ <---- negative, so it's a max.

Plug it back into the diagonal equation to get the height.
$60^2 = h^2 + (\sqrt{1200})^2$
$3600 = h^2 + 1200$
$2400 = h^2$
$h = \sqrt{2400}$

The dimensions should be $\sqrt{1200}\quad x \quad \sqrt{2400}$

How's this?

### optimization calculus beam strength

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