maximum area question (1 Viewer)

miracu · Nov 20, 2016

Screen Shot 2016-11-20 at 9.56.41 pm 10.47.49 PM.png

please help

android0 · Nov 21, 2016

$$Let the width of the rectangle be w, and its length, l. The expression for the perimeter is $P = 2w +l + \frac{\pi \times l}{2}=4.$

$Making w the subject of the equation: $w=2-\frac{(2+\pi)\times l}{4}. $ (1)$

$The area is given by A = $lw +\frac{\pi(\frac{l}{2})^2}{2}=\frac{\pi l^2}{8}+lw. $ (2)$

$Substituting (1) into (2): $A= \frac{l^2 (-\pi-4)}{8}+2l. $ (3)$

$This is a concave down parabola so finding the turning point will maximise the area.$

$$Differentiating, $ A'=\frac{(-\pi -4)l}{4} + 2 = 0$

$l=\frac{-8}{-\pi-4}= 1.12$

$$Substituting l = 1.12 into (3), $ A = 1.12m^2$

maximum area question (1 Viewer)

miracu

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android0

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