#### Provide solution for RD Sharma maths class12 Chapter Maxima and Minima exercise 17.5 question 8.

$\frac{80\sqrt{3}}{9+4\sqrt{3}}\; \text{and}\; \frac{180}{9+4\sqrt{3}}$

Hint: For maximum or minimum value of z must have $\frac{dz}{dx}=0$

Given: Suppose wire which has to made into squre and triangle is cut into two pieces x and y respectively.

Solution: x + y  = 20           ......(1)

Perimeter of square = 4a = x

$a=\frac{x}{4}$

Area of square $a^2=\left (\frac{x}{4} \right )^2=\frac{x^2}{16}$

Perimeter of triangle  3a=y

$a=\frac{y}{3}$

Area of triangle $=\frac{\sqrt{3}}{4}a^2$

$=\frac{\sqrt{3}}{4}\left (\frac{y}{3} \right )^2=\frac{\sqrt{3}y^2}{36}$

Now,

z =Area of square + area of triangle

\begin{aligned} &z=\frac{x^{2}}{16}+\frac{\sqrt{3} y^{2}}{36} \\ &=\frac{x^{2}}{16}+\frac{\sqrt{3}(20-x)^{2}}{36} \\ &\frac{d z}{d x}=\frac{2 x}{16}+\frac{2 \sqrt{3}(20-x)}{36} \end{aligned}

For minimum or maximum of z

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

\begin{aligned} &\frac{2 x}{16}+\frac{\sqrt{3}(20-x)}{18}=0 \\ &\frac{9 x}{4}=\sqrt{3}(20-x) \\ &\frac{9 x}{4}+x \sqrt{3}=20 \sqrt{3} \end{aligned}

\begin{aligned} &x\left(\frac{9}{4}+\sqrt{3}\right)=20 \sqrt{3} \\ &x=\frac{20 \sqrt{3}}{\left(\frac{9}{4}+\sqrt{3}\right)} \\ &x=\frac{80 \sqrt{3}}{(9+4 \sqrt{3})} \end{aligned}

$y=20-\frac{80 \sqrt{3}}{(9+4 \sqrt{3})}$

$y=20-\frac{180}{(9+4 \sqrt{3})}$                    from equation (1)

$\frac{d^{2} z}{d x^{2}}=\frac{1}{8}+\frac{\sqrt{3}}{18}>0$

Thus z is maximum when x $\frac{80 \sqrt{3}}{(9+4 \sqrt{3})}$