Originally Posted by

**drewms64** Thanks for the advice on both of the questions, for this one would it then end up being,

x=b y=l

$\displaystyle

A(y)= \left(\frac{98}{y-4}+2 \right)y

$

$\displaystyle

= \left(\frac{98y}{y-4}+2y \right)

$

$\displaystyle

A'(y)= \left(\frac{(y-4)(98)-(98y)(1)}{(y-4)^2)}+2 \right)

$

$\displaystyle

= \left(\frac{-392}{(y-4)^2}+2 \right) = 0

$

not sure how to get [tex] to work for this one, but

-392(y-4)^-2 = -2

(y-4)^-2 = 196

take the -2 root of each side

y-4 = - 14

y = -10

then when I find x I get -5

I dont think these should be negative though?