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

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

Answers (1)

 \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})}

 

Posted by

infoexpert24

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