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

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

## closed as off-topic by The Chaz 2.0, Daniel W. Farlow, Daniel, user91500, Claude LeiboviciMar 17 '17 at 7:59

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• You need to express the length, width and area of the rectangle in terms of $x$ and find the value of $x$ which gives the maximum area. – John Wayland Bales Mar 16 '17 at 18:59
• as far as i went , i considered rectangle with maximum area is square , if side length is L , so point B coordinate is B(L,32-L) the same as B(X,32-X) , it lies on the curve so it verifies it , i did the algebra and got a number , but Iam not sure , and the method is still so loosy – Mohamed Khaled Mar 16 '17 at 19:27

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 $f(2)=48$.

• great , but can i ask you if there is a way , using calculus , i hadnt studied am-gm , and its a calculus test , ik u already answered it , but i hope you can help me with that – Mohamed Khaled Mar 17 '17 at 12:33

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 $$A=a\cdot \left(32-a^3\right)$$

• any explanation , + its a test sample , and Iam supposed to give numerical result – Mohamed Khaled Mar 16 '17 at 19:22
• What you have left is to derive $A(a)$, then solve the equation $A'(a)=0$ ... – Bernard Massé Mar 16 '17 at 19:39
• A=a⋅(32−f(a)) where this function came from – Mohamed Khaled Mar 16 '17 at 20:29