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

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

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