Help with calculus question

a) [MATH]V=x^2y\implies y=\frac{V}{x^2}[/MATH]
b) [MATH]A=2x^2+4xy=2x^2+\frac{4V}{x}[/MATH]
Given that \(V=16\text{ cm}^2\) we then have:

[MATH]A=2x^2+\frac{64}{x}[/MATH]
c) [MATH]\d{A}{x}=4x-\frac{64}{x^2}=4\left(\frac{x^3-16}{x^2}\right)=0\implies x=2\sqrt[3]{2}[/MATH]
[MATH]A'(2)<0[/MATH]
[MATH]A'(3)>0[/MATH]
Thus, the first-derivative test confirms the critical value is at a local minimum.

d) [MATH]\frac{d^2A}{dx^2}=4+\frac{64}{x^3}[/MATH]
We see that the area function will be concave up at the critical value, thus by the second-derivative test we conclude the critical value is at a local minimum.

e) [MATH]A(2\sqrt[3]{2})=2(2\sqrt[3]{2})^2+\frac{64}{(2\sqrt[3]{2})}=24\cdot2^{\frac{2}{3}}[/MATH]
[MATH]y=\frac{16}{(2\sqrt[3]{2})^2}=2\sqrt[3]{2}=x[/MATH]
Since \(y=x\), what special type of cuboid to we have?
 
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