find two real numbers whose sum is 10 and whose product is a maximum. equate the answer.?

Answer 1

x + y = 10 => y = 10 - x

x*y = x*(10 - x) = 10x - x^2 = f(x)

optimum = when f'(x) = 0

f'(x) = 10 - 2x

10 - 2x = 0 => x = 5


So: when x = 5 and y = 5, you get the maximum product.

Answer 2

The answer you are looking is the answer to the maximum of the function f(x) = a*x^2 + b*x + c, where a =/= 0

x + y = 10 => y = 10 - x

f(x) = x*y = x * (10 - x) = -x^2 + 10 * x

Since a = -1 < 0, function f(x) has a maximum. Coordinates of maximum is: T( -b / (2*a), (4*a*c - b^2) / 4* a)

We have a = -1, b = 10, c = 0, so we can now get 'x' from point T, and then can easily get y, without derivations other people used to explain.

x = -b / (2*a) = -10 / 2*(-1) = -10 / -2 = 5.

y = 10 - x = 10 - 5 = 5.

This seems like easier solution for the problem since derivations are learned much later than function f(x) = ax^2+bx+c, at least here in europe.
Of course, you can use derivations as well, and get the result as f'(x)=10 - 2x, where f(x)=0, and that is for x = 5.

Answer 3

5 x 5

Answer 4

Others have given with calculus but i given with algebra for another method
let numbers be x and 10-x as sum is 10

x(10-x) = 10x-x^2
to make it perfect square add and subtract 25
= 25-(25-10x+x^2)
= 25-(x-5)^2
clearly maximum when x-5 = 0 or x = 5

Answer 5

Call one number x and then the other number is (10-x). If the product is to be a maximum, then let x(10-x) = y and find the point at which the derivative dy/dx = 0
dy/dx = 10-2x so at 10-2x = 0, x = 5.


Doug

Answer 6

the answer is 10.

Answer 7

x + y = 10
xy = max

ANS : x = 5 and y = 5, since you didn't say anything about the number not equalling each other.