Cfrgtkky

Having trouble with this optimization question and was hoping I could get some help with it. The function of the curve is $8^{-frac{x}{5}}$. I would greatly appreciate a full explanation.

I already have that: $A = W times H$, $A = 2x times 8^{-frac{x}{5}}$, $A = 16 x^{-frac{x}{5}}$.

I think the derivative of this is $displaystyle frac{-16log x - 16}{5} frac{x^x}{5}$, and in order to get the maximum I have to set it equal to $0$ and solve for it, but I'm somewhat stuck.

Yes, the area is $x 8^{-x/5}$. Take logarithms to obtain

$$log x - (xlog 8)/5,$$

which is maximized at $x=5/log 8$,

which (if I didn't screw things up) yields

$5/(elog 8)$.

Since $8^{-frac{x}{5}}>0$, for $xge 0$ the area is given by $$A(x)=xcdot 8^{-frac{x}{5}}$$

the derivative is given by

$$A(x)=xcdot 8^{-frac{x}{5}}=e^{log x-frac{x}{5}log8}$$

$$A'(x)=e^{log x-frac{x}{5}log8} cdot left(log x-frac{x}{5}log8right)'=8^{-x/5}left(frac1x-frac{1}{5}log8right)=0 implies frac1x-frac{1}{5}log8=0 implies x=frac{5}{log 8}\implies A_{MAX}=frac{5}{log 8}cdot 8^{-frac{1}{log 8}}=frac{5}{log 8}cdot 8^{-log_8e}=frac{5}{elog 8}$$