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Question
Find two positive numbers x and y such that x + y = 60 and xy3 is maximum.
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Solution
Let the two number be x and y and x + y = 60. ...(i)
Let P = xy3
⇒ P = (60 - y)y3 ...[from(i)]
⇒ P = 60y3 - y4
⇒ `(dP)/dy = 180y^2 - 4y^3`
For maximum P, we must have `(dP)/dy = 0`
⇒ 180y2 - 4y3 = 0
⇒ 4y2 (45 - y) = 0
⇒ y = 45 ...(∵ 0 < y < 60)
Also, `(d^2P)/dy^2 = 360y - 12y^2 and`
`((d^2P)/dy^2)_(y = 45) = 360 xx 45 - 12 xx (45)^2 < 0`
Therefore, P is maximum when y = 45
∴ The required numbers are x = 60 - y = 60 - 45 = 15 and y = 45.
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