# Thread: Optimizations problem with derivatives

1. ## Optimizations problem with derivatives

Question: (Attached) 59 b and c (I did a).

Answers:
b: About 8.5 inches by 2 inches.
c: 20/sqrt(3) inches by 20*sqrt(2/3) inches

My work is also attached for #59 a (Page 12/14 of pdf file) which I did successfully. I cannot complete b to get to c and for b my problem is that, I cannot find a relationship between the two variables to differentiate a function with one variable. As for #59 b, I thought I was on to something and wrote some unnecessary (I think) things.

Any help would be greatly appreciated!
Thanks in advance!

2. I'm not sure if I'm correct, but after several mistakes I derived an answer for the width of the plank slightly different from $\displaystyle 8.5\,\mbox{in}$. I used the equations

\displaystyle \begin{aligned} y&=\sqrt{100-x^2}-\sqrt{10}\\ A&=2xy=2x(\sqrt{100-x^2}-\sqrt{10}). \end{aligned}

Differentiating, I obtained

\displaystyle \begin{aligned} A'&=2\sqrt{100-x^2}+2x\cdot\frac{1}{2}(100-x^2)^{-\frac{1}{2}}\cdot(-2x)-2\sqrt{10}\\ &=2\sqrt{100-x^2}-2x^2(100-x^2)^{-\frac{1}{2}}-2\sqrt{10}=0\quad\mbox{at max}. \end{aligned}

Adding $\displaystyle 2\sqrt{10}$, multiplying both sides by $\displaystyle \sqrt{100-x^2}$, and squaring gives us a quadratic equation from which I derived my final answer.