Re: Two shapes and one line.

Quote:

Originally Posted by

**Nervous** The furthest I've gotten is

$\displaystyle 60 = 4x + 3y $

PS, this is just for a test I'm taking online, I'm not even in a calculus class. Don't think I'm trying to cheat or anything.

Well if you cut the wire at length x, then the remaining length is 60 - x.

If you make the square with the length 60 - x, then its side length is $\displaystyle \displaystyle \begin{align*} \frac{60 - x}{4} \end{align*}$ and its area is $\displaystyle \displaystyle \begin{align*} \left(\frac{60-x}{4}\right)^2 = \frac{3600 - 120x + x^2}{16} \end{align*}$

If the triangle is made with the length x, then its base is x and its height is $\displaystyle \displaystyle \begin{align*} \frac{\sqrt{3}\,x}{2} \end{align*}$, therefore its area is $\displaystyle \displaystyle \begin{align*} \frac{\sqrt{3}\,x^2}{4} \end{align*}$

So the total area is $\displaystyle \displaystyle \begin{align*} \frac{3600 - 120 + \left(4\sqrt{3} + 1\right)x^2}{16} \end{align*}$

Can you figure out how to find the maximum area now?

Re: Two shapes and one line.

I derived the fraction and got:

$\displaystyle \frac{-120+2x+8\sqrt{3}x^2}{16}$

I set it equal to zero and got:

$\displaystyle x=~7.568$

And when I plugged that into the original equation, I got ~196.620

But, the answer is supposed to be 225...

Re: Two shapes and one line.

You equated area to zero! You should equate its derivative to zero in order to obtain x value that makes area maximum $\displaystyle \frac{60}{1+4 \sqrt{3}}$ or something like that!

Re: Two shapes and one line.

$\displaystyle \frac{60}{1+4 \sqrt{3}} = ~7.568$