Find two positive numbers `x` and `y` such that their sum is 35 and the product `x^2y^5`is a maximum.

Explanation

Let ` P = x^(2)y^(5)`
Now ` x+y = 35 rArr x = 35 - y`
`:. P = x^(2)y^(5) = (35 - y)^(2) * y^(5)`
` rArr (dP)/(dy) = (35 - y)^(2) 5y^(4)+y^(5)2(35-y)(-1)`
`=y^(4)(35-y)[5(35-y)-2y]`
` = y^(4)(35-y)(175 - 5y-2y)`
` =y^(4)(35-y)(175-7y)`
` = (35y^(4)-y^(5))(175-7y)`
and ` (d^(2)P)/(dy^(2)) = (35y^(4)-y^(5)(-7)+(175-7y)(4 xx 35 xx y^(3)-5y^(4))`
` = - 7y^(4)(35-y)+7(25-y)xx 5y^(3)(28-y)`
` = - 7y^(4) (35-y)+35y^(3) (25 - y)(28-y)`
For maximum value, ` (dP)/(dy) = 0`
` rArr y^(4)(35-y)(175-7y)=0`
` rArr y=0, y = 25, y = 35`
when ` y=0, = 35 - 0 = 35" then the product "x^(2)y^(5) = 0`
when ` y= 35, x = 35 - 35 = 0" then the product "x^(2)y^(5) = 0`
`:. ` y = 0 and y = 35 are impossible.
when y = 25 then
`((d^(2)P)/(dy^(2)))_(y=25) = - 7 xx (25)^(4) xx (35 - 25) + 35 xx (25)^(3)`
` xx (25 - 25)(28 - 25)`
` =- 7 xx 390625 xx 10 +0 lt 0`
`:. ` P in maximum.
Thus, ` y = 25 and x = 35 - 25 = 10`
So the number are 10 and 25.