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  • Please solve RD Sharma class 12 Chapter maxima and minima exercise 17.5 question 4 maths textbook solution.

Please solve RD Sharma class 12 Chapter maxima and minima exercise 17.5 question 4 maths textbook solution.

Answers (1)

x = 6 and y =9

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

Given:  x + y =15

Solution: Let the two numbers be x any y then,

x + y = 15                          .......(1)

Now,

z = x^2y^3

z = x^2(15-x)^3                     from (1)

\frac{d z}{d x}=2 x(15-x)^{3}-3 x^{2}(15-x)^{2}

For minimum and maximum value of z

\begin{aligned} &\frac{d z}{d x}=0 \\ &2 x(15-x)^{3}-3 x^{2}(15-x)^{2}=0 \\ &2 x(15-x)=3 x^{2} \\ &30 x-2 x^{2}=3 x^{2} \\ &30 x=5 x^{2} \\ &x=6 \text { and } y=9 \end{aligned}

\frac{d^{2} z}{d x^{2}}=2(15-x)^{3}-6 x(15-x)^{2}-6 x(15-x)^{2}+6 x^{2}(15-x)

At  x =6

\begin{aligned} &\frac{d^{2} z}{d x^{2}}=2(9)^{3}-36(9)^{2}-36(9)^{2}+6(36)(9) \\ &\frac{d^{2} z}{d x^{2}}=-2430<1 \end{aligned}

Thus z is maximum when x =6 and y =9

 

 

Posted by

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