A square piece of steel, s cm on a side, is to made into an equipment chassis by cutting equal squares out of the corners, folding up the sides , and welding the seam to form a pan.
$A)$ If $s=60$, what should the side of the cutout square be, to maximize the volume of the chassis?
$B)$ In general, what is the relationship between the side of the steel square s and the side of the cutout square x that will produce the maximum volume of the chassis.
$A-$ So I'm not sure about this but I know that $V=s^3$ so $V=x(60-2x)(60-2x)$ then $V= 4x^3 -240x^2 + 3600x$ therefore, $V'= 12x^2 - 480x +3600$ so it's $V = 12(x-10)(x-30)$ $x= 10, x= 30$ Now I don't know what to use, I was thinking $x=30$ to maximize the volume and end up with $V=0$ so I'll have $S=15$ since it's 4 sides, but I read the question and $folding$ $up$ $the$ $sides$, so I should use $x=10$ ?but I'll get $V= 10[60-2(10)][60-2(10)]$ which is $16,000$ I'm confused. I also know that $V''= 24x-480$ so $x=10(Maxima)$ $x=30(Minima)$ using the second derivative test.
$B-$ I'm confused.
For part A, your answer of $x=10$ is correct. As you have shown, this gives a local maximum by the second derivative test, and the other critical point of $x=30$ does not make sense in the problem since then two sides of the pan would have length 0.
For part B, using the same method as in the first part gives $V=x(s-2x)^2$ for $0<x<\frac{s}{2}$, so $V^{\prime}(x)=x\cdot2(s-2x)(-2)+(s-2x)^2=(s-2x)(s-6x)=0 \iff x=\frac{s}{6}$.
By the first-derivative test, $x=\frac{s}{6}$ gives a maximum.
(Notice that $x=\frac{s}{2}$ does not make sense in the problem since then $s-2x=0$.)