Let x,y,z be non-negative real numbers satisfying the condition x+y+z=1 . The maximum possible value of x^3*y^3+y^3*z^3+z^3*x^3 has the form a/b where a and b are positive, coprime integers. What is the value of a+b ?
To find the maximum possible value of a given expression, we can use calculus. Let's consider the function f(x, y, z) = x^3*y^3 + y^3*z^3 + z^3*x^3.
First, we need to impose the constraint x + y + z = 1. To simplify things, we can rewrite this constraint as z = 1 - x - y.
Now, substitute z in terms of x and y into the function f(x, y, z):
f(x, y) = x^3 * y^3 + y^3 * (1 - x - y)^3 + (1 - x - y)^3 * x^3.
To find the maximum of this function, we need to take partial derivatives with respect to x and y and set them equal to zero.
∂f/∂x = 3x^2 * y^3 - 3y^3 * (1 - x - y)^2 + 3(1 - x - y)^2 * x^2 = 0.
∂f/∂y = 3x^3 * y^2 - 3(1 - x - y)^2 * y^2 + 3(1 - x - y)^2 * x^3 = 0.
Simplifying these equations, we get:
x^2 * y^2 - (1 - x - y)^2 * y^2 + (1 - x - y)^2 * x^2 = 0.
x^3 * y - (1 - x - y)^2 * y + (1 - x - y)^2 * x^3 = 0.
Solving these equations is a bit tedious, but we can use numerical methods or software to find the critical points. In this case, the critical points are x = y = z = 1/3.
Now, we need to check the endpoints of the feasible region. Since x, y, and z are non-negative, the endpoints are (1, 0, 0), (0, 1, 0), and (0, 0, 1).
Evaluating the function f at these critical points and endpoints:
f(1/3, 1/3, 1/3) = (1/3)^6 = 1/729.
f(1, 0, 0) = f(0, 1, 0) = f(0, 0, 1) = 0.
So, the maximum value of f is 1/729.
The expression 1/729 can be written as 1/729 = 1/3^6 = 1/(3^2)^3. Since 3 and 2 are coprime, the value of a (1) and b (729) share no common factors other than 1.
Therefore, the value of a + b is 1 + 729 = 730.