Find the maximum and minimum values?

Question

Find the maximum and minimum values of the function f(x, y)=xy^2 subject to the constraint x^2+y^2=4.

gradient of f=<y^2, 2xy>

gradient of g=<2x, 2y>

I'm stucked.

Answer 1

First of all, since the constraint's graph is a circle, which is a closed loop, and f is continuous in R², there must exist a constrained global maximum and minimum.

Lagrange Multipliers:

f(x,y) = xy²

g(x,y) = x² + y²

We want ∇f = λ∇g so we get the following system of equations

1. y² = 2λx

2. 2xy = 2λy

3. x² + y² = 4 ← The constraint.

Equation 2 implies that y = 0 or λ = x

We can ignore y = 0, since that will make f(x,y) = 0 and clearly f(x,y) takes on both positive and negative values subject to the constraint.

Plugging in the alternative, λ = x to equation 1 gives y² = 2x².

Plugging this into the constraint gives 3x² = 4 so that x² = 4/3 and y² = 8/3

Taking square roots gives

x = ±√(4/3) = ±(2/3)√3

y = ±√(8/3) = ±(2/3)√6

At the points < (2/3)√3 , ±(2/3)√6 >, f(x,y) = (16/9)√3 ← Maximum

At the points < -(2/3)√3 , ±(2/3)√6 >, f(x,y) = -(16/9)√3 ← Minimum

Answer 2

Well, the constraint makes this problem pretty easy. The constraint says that the only values of x and y that are allowed in this problem must obey the equation x^2 + y^2 = 4. This also means that y^2 = 4 - x^2. Since y^2 = 4 - x^2 let us substitute 4 - x^2 for y^2 in f(x, y) giving us: f(x, y) = x* (4 - x^2) notice how the value of f does not depend on y anymore. This means that we have simplified the problem to a one variable problem giving us: f(x) = x * (4 - x^2) f(x) = 4x - x^3 Now hopefully you will be able to find the maximum and minimum of this problem, but I'll step through it anyway, and there's a bit of a trick at the end. So, taking the first derivative of this problem to find the critical points we get: f'(x) = 4 - 3x^2 finding zeroes: 0 = 4 - 3x^2 = 3x^2 - 4 3x^2 = 4 x^2 = 4/3 x =(+ or -) 2 / sqrt (3) x = (+ or -) 2 sqrt (3) / 3 so the critical points are x = 2 sqrt(3) / 3, or -2 sqrt(3) / 3 Now let's just check if these are maximums and minimums by taking the second derivative: f'(x) = 4 - 3x^2 f''(x) = -6x f''(2 sqrt(3) / 3) = -6 * (2 sqrt(3) / 3) = -4 sqrt(3) the second derivative of the first critical point is negative which means curve at that point is concave down and so the point is a maximum, and you will find that the second derivative of the second critical point is positive which means the curve is concave up and so the point must be a minimum. Now in order to find the point in the x, y plane we need to look at the constraints. We know that x^2 = 4/3 (from when we were finding the critical points) so since x^2 + y^2 = 4 then 4/3 + y^2 = 4 and y^2 = 8/3 and y = 2 sqrt(6) / 3 So the two critical points are: (2 sqrt(3) / 3, 2 sqrt(6) / 3) and (-2 sqrt(3) / 3, 2 sqrt(6) / 3) Hope that helps!