1. ## More Optimization Problems

1. Find the minimum value of A if A=4y + x^2, where (x^2 + 1)y=324.

2. By cutting away identical squares from each corner of a rectangular piece of cardboard and folding up the resulting flaps, the cardboard may be turned into an open box. If the cardboard is 16 inches long and 10 inches wide, find the dimensions of the box that will yield maximum volume

2. ## #1

solving for y

$y=\frac{324}{x^2+1}$

and

$A=4y+x^2=4\cdot \frac{324}{x^2+1}=\frac{1296}{x^2+1}+x^2$

$A=\frac{1296}{x^2+1}+x^2$ taking the derivative

$\frac{dA}{dx}=\frac{-1296}{(x^2+1)^2}(2x)+2x=2x\left( 1-\frac{(36)^2}{(x^2+1)^2}\right)=2x \left( \frac{(x^2+1)^2-(36)^2}{(x^2+1)^2}\right)$

$=2x \left( \frac{(x^2+1-36)(x^2+1+36)}{(x^2+1)^2}\right)=
2x \left( \frac{(x^2-35)(x^2+37)}{(x^2+1)^2}\right)
$

Finally setting equal to we get three critical values

$x=0,x= \pm \sqrt{35}$

so now we get three y values

$y=\frac{324}{x^2+1}$

$(0,324) \mbox{ and } (\pm \sqrt{35},9)$

$A(0,324)=1296$ and $A(\pm \sqrt{35},9)=71$

so our min is at $(\pm \sqrt{35},9)$

3. for the last one minimize the volume

$V=(16-2x)(10-x)x$

Good luck.