PUC Karnataka Science Class 12Department of Pre-University Education, Karnataka
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# A Large Window Has the Shape of a Rectangle Surmounted by an Equilateral Triangle. If the Perimeter of the Window is 12 Metres Find the Dimensions of the Rectangle Will - PUC Karnataka Science Class 12 - Mathematics

#### Question

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

#### Solution

$\text { Let the dimensions of the rectangle be x and y } .$

$\text { Perimeter of the window = x + y + x + x + y = 12 }$

$\Rightarrow 3x + 2y = 12$

$\Rightarrow y = \frac{12 - 3x}{2} ...........\left( 1 \right)$

$\text { Area of the window } =xy+\frac{\sqrt{3}}{4} x^2$

$\Rightarrow A = x\left( \frac{12 - 3x}{2} \right) + \frac{\sqrt{3}}{4} x^2$

$\Rightarrow A = 6x - \frac{3 x^2}{2} + \frac{\sqrt{3}}{4} x^2$

$\Rightarrow \frac{dA}{dx} = 6 - \frac{6x}{2} + \frac{2\sqrt{3}}{4}x$

$\Rightarrow \frac{dA}{dx} = 6 - 3x + \frac{\sqrt{3}}{2}x$

$\Rightarrow \frac{dA}{dx} = 6 - x\left( 3 - \frac{\sqrt{3}}{2} \right)$

$\text { For maximum or a minimum values of A, we must have }$

$\frac{dA}{dx} = 0$

$\Rightarrow 6 = x\left( 3 - \frac{\sqrt{3}}{2} \right)$

$\Rightarrow x = \frac{12}{6 - \sqrt{3}}$

$\text { Substituting the value of x in eq. } \left( 1 \right), \text { we get }$

$y = \frac{12 - 3\left( \frac{12}{6 - \sqrt{3}} \right)}{2}$

$\Rightarrow y = \frac{18 - 6\sqrt{3}}{6 - \sqrt{3}}$

$\text { Now, }$

$\frac{d^2 A}{d x^2} = - 3 + \frac{\sqrt{3}}{2} < 0$

$\text { Thus, the area is maximum when }x=\frac{12}{6 - \sqrt{3}}\text { and }y=\frac{18 - 6\sqrt{3}}{6 - \sqrt{3}}.$

#### Notes

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Solution A Large Window Has the Shape of a Rectangle Surmounted by an Equilateral Triangle. If the Perimeter of the Window is 12 Metres Find the Dimensions of the Rectangle Will Concept: Graph of Maxima and Minima.
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