A box open from the top is made from a rectan | vedclass

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Question

(C) The dimensions of the box formed are $(a-2x)$,$(b-2x)$,and $x$. The volume $V$ of the box is given by: $V(x) = (a-2x)(b-2x)x = (ab - 2ax - 2bx + 4

Vedclass · Accepted Answer

$\frac{a+b-\sqrt{a^{2}+b^{2}-ab}}{6}$

Vedclass · Answer

(C) The dimensions of the box formed are $(a-2x)$,$(b-2x)$,and $x$. The volume $V$ of the box is given by: $V(x) = (a-2x)(b-2x)x = (ab - 2ax - 2bx + 4x^2)x = 4x^3 - 2(a+b)x^2 + abx$. To find the maximum volume,we differentiate $V(x)$ with respect to $x$: $\frac{dV}{dx} = 12x^2 - 4(a+b)x + ab$. Setting $\frac{dV}{dx} = 0$: $12x^2 - 4(a+b)x + ab = 0$. Using the quadratic formula $x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}$: $x = \frac{4(a+b) \pm \sqrt{16(a+b)^2 - 48ab}}{24} = \frac{4(a+b) \pm \sqrt{16(a^2 + 2ab + b^2 - 3ab)}}{24} = \frac{4(a+b) \pm 4\sqrt{a^2 - ab + b^2}}{24} = \frac{(a+b) \pm \sqrt{a^2 - ab + b^2}}{6}$. Let $\alpha = \frac{(a+b) + \sqrt{a^2 - ab + b^2}}{6}$ and $\beta = \frac{(a+b) - \sqrt{a^2 - ab + b^2}}{6}$. Using the second derivative test: $\frac{d^2V}{dx^2} = 24x - 4(a+b)$. For $x = \beta$,$\frac{d^2V}{dx^2} = 24\left(\frac{a+b - \sqrt{a^2 - ab + b^2}}{6}\right) - 4(a+b) = 4(a+b) - 4\sqrt{a^2 - ab + b^2} - 4(a+b) = -4\sqrt{a^2 - ab + b^2} Since the second derivative is negative at $x = \beta$,the volume is maximum at $x = \beta = \frac{a+b - \sqrt{a^2 - ab + b^2}}{6}$.

$A$ box open from the top is made from a rectangular sheet of dimensions $a \times b$ by cutting squares of side $x$ from each of the four corners and folding up the flaps. If the volume of the box is maximum,then $x$ is equal to:

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