Question

Find two positive numbers x and y such that their sum is 35 and the product x²y⁵ is a maximum.

Answer 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

Answer 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.