Find the maximum area of rectangle between y-axis , $f(x)=x^3 , y=32$

Question

Find the maximum area of rectangle between the $y$-axis , $f(x)=x^3 , y=32$

Answer 1

Let $B(t,t^3)$, where $t>0$ and $t^3<32$.

Thus, $AB=t$ and $BC=32-t^3$.

Hence, by AM-GM we obtain: $$S_{ABCD}=t(32-t^3)=48-(t^4-32t+48)=$$ $$=48-(t^4+16+16+16-32t)\leq48-\left(4\sqrt[4]{t^4\cdot16^3}-32t\right)=48.$$ The equality occurs for $t=2$.

Id est, the answer is $48$.

By calculus it's harder:

Let $f(t)=32t-t^4$, where $0<t<\sqrt[3]{32}$.

Hence, $f'(t)=32-4t^3=4(2-t)(4+2t+t^2)$.

Since $f'(t)>0$ for $0<t<2$ and $f'(t)<0$ for $2<t<\sqrt[3]{32}$,

we obtain $t_{max}=2$ and the answer is $f(2)=48$.

Answer 2

let one sidelength of rectangle be $a$ then we get the area as $$A=a\cdot (32-f(a))$$ if $$f(x)=x^3$$ then we have $$A=a\cdot \left(32-a^3\right)$$