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#### Provide solution for RD Sharma maths class12 Chapter Maxima and Minima exercise 17.5 question 7.

$\frac{112}{\pi +4}\; \text{and}\; \frac{28 \pi }{\pi +4}$

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

Given: Suppose the wire which is to be made into a square and circle is cut into two pieces of length x,m and ym respectively.

Solution: x +y = 28                  ......(1)

Perimeter of square 4a=x

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

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

Circumference of circle $2\pi r=y$

$r=\frac{y}{2\pi}$

Area of circle $=\pi r^2$

$=\pi \left ( \frac{y}{2\pi } \right )^2=\frac{y^2}{4\pi }$

z = Area of square + Area of circle

\begin{aligned} &=\frac{x^{2}}{16}+\frac{y^{2}}{4 \pi} \\ &=\frac{x^{2}}{16}+\frac{(28-x)^{2}}{4 \pi} \\ &\frac{d z}{d x}=\frac{2 x}{16}-\frac{2(28-x)}{4 \pi} \end{aligned}

For maximum or minimum value of z

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

\begin{aligned} &\frac{2 x}{16}-\frac{2(28-x)}{4 \pi}=0 \\ &\frac{x}{4}=\frac{(28-x)}{\pi} \\ &\frac{x \pi}{4}+x=28, x\left(\frac{\pi}{4}+1\right)=28 \end{aligned}

\begin{aligned} x &=\frac{28}{\left(\frac{\pi}{4}+1\right)} \\ x &=\frac{112}{\pi+4} \end{aligned}

$y=28-\frac{112}{\pi +4}$                from equation 1

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

Thus z is maximum when $\frac{112}{\pi +4}$ and $\frac{28 \pi }{\pi +4}$