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Question
Find two positive numbers x and y such that their sum is 35 and the product x2y5 is a maximum.
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Solution 1
Let the number be x and y and let p = x2y2 and
x + y = 35 ... (i)
⇒ p = (35 − y)2y2 ... [from (i)]
Now, `(dp)/dy = (35 - y)^2 (5y^4) + y^5 xx 2 (35 - y) (-1)`
y4 (35 − y) {5 (35 − y) − 2y}
= y4 (35 − y) (175 − 7y)
For maximum p, let `(dp)/dy = 0`
⇒ y4 (35 − y) (175 − 7y) = 0
⇒ 175 − 7y = 0 ...(∵ 0 < y < 35)
⇒ y = 25
Now,
`((d^2p)/dy^2) = 4 (35 - y) (175 - 7y)y^3 + y^4 (-1) (175 - 7y) + y^4 (35 - y) (-7)`
⇒ `((d^2p)/dy^2)_(y = 25) < 0`
and p has a maximum value at y = 25
∴ The required numbers are x = 35 − 25 = 10 and y = 25
Solution 2
x, y > 0 with x + y = 35. Maximize f = x2y5
max ln f = 2 ln x + 5 ln y s.t. x + y = 35.
`∂/(∂x): 2/x = lambda, ∂/(∂y) : 5/y = lambda => 2/x = 5/y => x/y = 2/5`
With x + y = 35:
`x = 2/7 xx 35 = 10, y = 5/7 xx 35 = 25`
So the product is maximized at x = 10, y = 25
x2y5 = 102 ⋅ 255 = 976,562,500.
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