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Question Number 40479 by Raj Singh last updated on 22/Jul/18

Answered by MJS last updated on 22/Jul/18

$$a=8+2x$$$$b=c=d=8$$$$h=\sqrt{64-x^2}$$$$A\left(x\right)=\frac{1}{2}(a+c)h=(8+x)\sqrt{64-x^2}$$$$A^{\prime }\left(x\right)=-2\frac{x^2+4x-32}{\sqrt{64-x^2}}$$$$A^{″}\left(x\right)=2\frac{x^2-8x-32}{(8-x)\sqrt{64-x^2}}$$$$A^{\prime }\left(x\right)=0\Rightarrow x=4$$$$A^{″}\left(4\right)=-2\sqrt{3}<0\Rightarrow \mathrm{maximum}\mathrm{at}x=4$$$$A\left(4\right)=48\sqrt{3}\approx 83.14{cm}^2$$

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