A square piece of tin of side 30,cm is to be | vedclass

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(C) Let the side of the square cut from each corner be $x\,cm$.The dimensions of the resulting box will be length $= (30-2x)\,cm$,width $= (30-2x)\,cm

Vedclass · Accepted Answer

$800$

Vedclass · Answer

(C) Let the side of the square cut from each corner be $x\,cm$.The dimensions of the resulting box will be length $= (30-2x)\,cm$,width $= (30-2x)\,cm$,and height $= x\,cm$.The volume $V$ of the box is given by $V = x(30-2x)^2$.To find the maximum volume,we differentiate $V$ with respect to $x$:$V = x(900 - 120x + 4x^2) = 4x^3 - 120x^2 + 900x$.$\frac{dV}{dx} = 12x^2 - 240x + 900$.Setting $\frac{dV}{dx} = 0$ for critical points:$12(x^2 - 20x + 75) = 0$$12(x-5)(x-15) = 0$.Thus,$x = 5$ or $x = 15$. Since $x=15$ would result in a side length of $0$,we take $x = 5\,cm$.The surface area $S$ of the open box is the area of the original square minus the area of the four cut-out squares:$S = (30)^2 - 4x^2 = 900 - 4(5)^2 = 900 - 100 = 800\,cm^2$.

$A$ square piece of tin of side $30\,cm$ is to be made into a box without a top by cutting a square of side $x$ from each corner and folding up the flaps to form a box. If the volume of the box is maximum,then its surface area (in $cm^2$) is equal to $............$.

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