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  • Need solution for RD Sharma maths class 12 chapter Maxima and Minima exercise 17.5 question 16.

Need solution for RD Sharma maths class 12 chapter Maxima and Minima exercise 17.5 question 16.

Answers (1)

x = \frac{12}{6-\sqrt{3}} \; \text{and}\; \: y = \frac{18-6\sqrt{3}}{6-\sqrt{3}}

 

Hint: For maxima and minima value of A, we must have \frac{dA}{dx}=0

Given:  3x + 2y = 12

Solution:

Let the dimensions of the rectangular part x, y

Perimeter of the window ⇒ x + y + x + x + y = 12

3x + 2y = 12

y = \frac{12-3x}{2}       .....(1)

Area = xy +\frac{ \sqrt{3}}{4}x^2

\begin{aligned} &A=x\left(\frac{12-3 x}{2}\right)+\frac{\sqrt{3}}{4} x^{2} \\ &=6 x-\frac{3 x^{2}}{2}+\frac{\sqrt{3}}{4} x^{2} \\ &\frac{d A}{d x}=6-\frac{6 x}{2}+\frac{2 \sqrt{3}}{4} x \\ &\frac{d A}{d x}=6-3 x+\frac{\sqrt{3}}{2} x \\ &=6-x\left(3-\frac{\sqrt{3}}{2}\right) \end{aligned}

\frac{dA}{dx}=0

\begin{aligned} &6=x\left(3-\frac{\sqrt{3}}{2}\right) \\ &x=\frac{12}{6-\sqrt{3}} \end{aligned}

Substitute x value in eqn (1)

\begin{aligned} &y=\frac{12-3\left(\frac{12}{6-\sqrt{3}}\right)}{2}=\frac{18-6 \sqrt{3}}{6-\sqrt{3}} \\ &\frac{d^{2} A}{d x^{2}}=-3+\frac{\sqrt{3}}{2}<0 \end{aligned}

Thus the area is max when x = \frac{12}{6-\sqrt{3}} \; \text{and}\; \: y = \frac{18-6\sqrt{3}}{6-\sqrt{3}}

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