I am a Calculus one student There are a bunch of these questions about finding the maximum volume of a box on Stack Exchange. I was looking at this question; Finding dimensions of a rectangular box
I was looking at answer number two and trying to see if I could follow, and instead of $108 cm^2$ replace it with A. The first equation I have to work on then is the surface area equation; $$A=xy+2xz+2yz$$ $$A=xy+2z(x+y)$$ $$z=\frac{A-xy}{2(x+y}$$
Then I can substitute the surface area constraint equation into the volume equation. $$Volume= xy \left(\frac{A-xy}{2(x+y)}\right)$$
Then I need to find $\frac{d}{dx} xy \left(\frac{A-xy}{2(x+y)}\right)$ and $\frac{d}{dy} xy \left(\frac{A-xy}{2(x+y)}\right)$
Here are the derivatives $$\frac{d}{dx} xy \left(\frac{A-xy}{2(x+y)}\right)=\frac{y^2(a-x^2-2xy)}{2(x^2+y^2+2xy)}$$ $$\frac{d}{dy} xy \left(\frac{A-xy}{2(x+y)}\right) = \frac{x^2(a-y^2-2xy)}{2(x^2+y^2+2xy)}$$
Then I set the derivative equal to zero for each equation.
The two equations I came up with are $$x=-\frac{-a+y^2}{2y}$$ $$y=-\frac{-a+x^2}{2x}$$
I am not really sure how to interpret this. I was looking to find out if I am doing something wrong and if not what the next step would be. It could be I plug these two equations into the constraint equation and find z. But, at this point I am pretty turned around and was looking for some advice and help. thank you
