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Exercise: Find rectangular box with surface area 10 that has minimum volume

My solution: We know that surface area is $$S= 2(wl +hl+hw)$$and volume $$V=whl$$ ,where w-width,h-heigh, l=length.

We will use lagrange function$$L = whl - \lambda(2(wl +hl+hw)-10)$$. Now we will find partial derivatives w.r.t $w,h,l,\lambda$ respectively $$1)hl-2h\lambda-2l\lambda \\ 2)wl-2l\lambda - 2wl\lambda \\ 3)wh-2w\lambda-2h\lambda \\ 4) 2(wl +hl+hw) - 10$$

By setting this equations to 0 we will get that $l=h=w$. Then inset into Surface equation and we will get that the volume will be the biggest when $w=h=l= \sqrt{\frac{5}{3}}$.

Did i solve it correctly?

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As an alternative and to check let use AM-GM inequality

$$\frac {S} 6= \frac{wl +hl+hw}{3}\ge\sqrt[3]{w^2l^2h^2}\implies V^{\frac23}\le\frac {5}3\implies V\le\sqrt{\left(\frac{5}3\right)^3}$$

with equality for $$wl=hl=hw \implies w=l=h=\sqrt{\frac{5}3}$$

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