Question

# Find two positive numbers x and y such that x + y = 60 and xy 3 is maximum.

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Solution

## The two numbers are given such that, x+y=60 It can be written as, y=60−x As x y 3 is given to be maximum, so substitute the value of y in the function, f( x )=x ( 60−x ) 3 Differentiate the function with respect to x, f ′ ( x )= d[ x ( 60−x ) 3 ] dx = ( 60−x ) 3 ×1+x×3 ( 60−x ) 2 ×( −1 ) = ( 60−x ) 2 ( 60−x−3x ) = ( 60−x ) 2 ( 60−4x ) (1) This gives x=60 or x=15. Differentiate equation (1) with respect to x, f ″ ( x )=−2( 60−x )( 60−4x )−4 ( 60−x ) 2 =−2( 60−x )[ 60−4x+2( 60−x ) ] =−2( 60−x )( 180−6x ) =−12( 60−x )( 30−x ) When x=60, f ″ ( x )=0 When x=15, f ″ ( x )=−12( 60−15 )( 30−15 ) =−12( 45 )( 15 ) <0 This shows that the point of local maxima is x=15, so, y=60−x =60−15 =45 Therefore, x=15 and y=45.

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