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# A window has the shape of a rectangle surmounted by an equilateral triangle. If the perimeter of the window is 12 m, find the dimensions of the rectangle that will produce the largest area of the window.

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Step 1:
Perimeter of the window is $P=3x+2y=12$
$2y=\large\frac{12-3x}{2}$
Area of the window =area of the rectangle+area of the triangle
$A=xy+\large\frac{\sqrt 3}{4}$$x^2 \;\;=x(\large\frac{12-3x}{2})+\frac{\sqrt 3}{4}$$x^2$
$\;\;=\large\frac{12x}{2}-\frac{3x^2}{2}+\frac{\sqrt 3}{4}$$x^2 A=6x-\large\frac{3}{2}$$x^2+\large\frac{\sqrt 3}{4}$$x^2 Step 2: \large\frac{dA}{dx}$$=6-\large\frac{6x}{2}+\frac{\sqrt 3\times 2x}{4}$
$\quad=6-3x+\large\frac{\sqrt 3}{2}$$x For the area to be maximum or minimum \large\frac{dA}{dx}$$=0$
$x(\large\frac{\sqrt 3}{2}$$-3)+6=0 \Rightarrow x(3-\large\frac{\sqrt 3}{2})=$$6$
$x=\large\frac{6\times 2}{6-\sqrt 3}=\frac{12}{6-\sqrt 3}$
$x=\large\frac{12(6+\sqrt 3)}{3}$
$x=4(6+\sqrt 3)$
Step 3:
Diff again w.r.t $x$ we get
$\large\frac{d^2A}{dx^2}=$$-3+\large\frac{\sqrt 3}{2}$$< 0$
Hence the dimension of the rectangle produce maximum area when
$x=4(6+\sqrt 3)$ and $y=12-\large\frac{3(6+\sqrt 3)}{2}$
$y=\large\frac{24-18-3\sqrt 3}{2}$
$\;\;=6-3\sqrt 3$
$y=3(2-\sqrt 3)$
Hence $x=4(6+\sqrt 3)$ and $y=3(2-\sqrt 3)$
answered Nov 11, 2013