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.