A rectangular sheet of tin 45 , cm by 24 , cm | vedclass

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(A) Let the side of the square to be cut be $x \, cm$. Then,the height of the box is $x$,the length is $45-2x$,and the breadth is $24-2x$. Therefore,t

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$5$

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(A) Let the side of the square to be cut be $x \, cm$. Then,the height of the box is $x$,the length is $45-2x$,and the breadth is $24-2x$. Therefore,the volume $V(x)$ of the box is given by: $V(x) = x(45-2x)(24-2x) = x(1080 - 90x - 48x + 4x^2) = 4x^3 - 138x^2 + 1080x$. Now,find the first derivative: $V'(x) = 12x^2 - 276x + 1080 = 12(x^2 - 23x + 90) = 12(x-18)(x-5)$. Setting $V'(x) = 0$,we get $x = 18$ or $x = 5$. Since the breadth is $24 \, cm$,$x$ cannot be $18$ (as $2x$ would be $36 > 24$). Thus,$x = 5$. Now,find the second derivative: $V''(x) = 24x - 276$. At $x = 5$,$V''(5) = 24(5) - 276 = 120 - 276 = -156 Since $V''(5) < 0$,the volume is maximum at $x = 5 \, cm$.

$A$ rectangular sheet of tin $45 \, cm$ by $24 \, cm$ is to be made into a box without a top by cutting off a square from each corner and folding up the flaps. What should be the side of the square to be cut off so that the volume of the box is maximum (in $, cm$)?

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