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(C) Let the side of the square base be $x$ and the height of the box be $y$.Given that the surface area of the box with an open top is $S = x^2 + 4xy

Vedclass · Accepted Answer

$32$

Vedclass · Answer

(C) Let the side of the square base be $x$ and the height of the box be $y$.Given that the surface area of the box with an open top is $S = x^2 + 4xy = 48$.From this,we can express $y$ in terms of $x$: $4xy = 48 - x^2 \Rightarrow y = \frac{48 - x^2}{4x}$.The volume of the box is $V = x^2y$.Substituting the value of $y$,we get $V = x^2 \left( \frac{48 - x^2}{4x} ight) = \frac{1}{4}(48x - x^3)$.To find the maximum volume,we differentiate $V$ with respect to $x$: $\frac{dV}{dx} = \frac{1}{4}(48 - 3x^2)$.Setting $\frac{dV}{dx} = 0$,we get $48 - 3x^2 = 0 \Rightarrow x^2 = 16 \Rightarrow x = 4$ (since $x > 0$).Now,find $y$ using $x = 4$: $y = \frac{48 - 4^2}{4(4)} = \frac{48 - 16}{16} = \frac{32}{16} = 2$.The maximum volume is $V = x^2y = 4^2 	imes 2 = 16 	imes 2 = 32 \, m^3$.

$A$ box is to be made with a square base and an open top. If the area of the material used is $48 \, m^2$,then the maximum volume of the box is ........... $m^3$.

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