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  • Need solution for RD Sharma maths class 12 chapter Maxima and Minima exercise 17.5 question 44.

Need solution for RD Sharma maths class 12 chapter Maxima and Minima exercise 17.5 question 44.

Answers (1)

x =15, y = 10

Hint: For maxima and minima value of S, we must have \frac{dS}{dD}=0

Given:

Let x and y be the length and breadth of the rectangle

Area of page = 150

xy = 150

y = \frac{150}{x}             ......(1)

Solution:

Area of printed matter = (x-3)(y-2)

A=xy-2x-3y+6

A=150-2x-\frac{450}{x}+6

\frac{dA}{dx}=-2+\frac{450}{x^2}

\frac{dA}{dx}=0

-2+\frac{450}{x^2}=0

2x^2=450

x=15

Sub x=15 value in eqn (1)

y = 10

Now,

\frac{d^2A}{dx^2}=\frac{-900}{x^3}=\frac{-900}{(15)^3}

=\frac{-900}{3375}<0

Thus area of printed matter is max when x = 15 and y = 10

Posted by

infoexpert24

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