f(x,y)= xy(3-x-y) What are the absolute maximum points of the function in the R region? What are the absolute minimum points of the function in the R region?

Question

R:x+y=4 between the line and the coordinate axes region.

Answer 1

To find the abesolute maximum/minimum points of a function in a region, you need to find the local maximum/minimum points of the function, find the local maximum/minimum points of the function along the boundaries, and compare them. To find the local maximum/minimum points of a function, you need to find the critical points of the function first.

First, let us find the critical points, namely solve ∂f/∂x=0 and ∂f/∂y=0. I assume you know how to find the partial derivative. So we will have a system of equations:

∂f/∂x = y(3-x-y)-xy = 0 (*) and ∂f/∂y = x(3-x-y)-xy = 0.

Subtract these two equations, we have

(x-y)(3-x-y)=0.

We have two cases: 1) x=y, 2) x+y=3. In the first case, substitute x=y into the equation (*), we have 3x-3x² =0, which gives us two solutions (0,0) and (1,1). In the second case, substitute x+y=3 into the equation (*), we have xy=0, which, together with x+y=3, gives us another two solutions (3,0) and (0,3).

So the critical points are (0,0), (1,1), (3,0), and (0,3). To decide whether a critical points are local maximum/minimum or not, you need to compute the determinant of the Hessian matrix

⌈∂²f/∂x² ∂²f/∂x∂y⌉

⌊∂²f/∂x∂y ∂²f/∂y²⌋.

The critical point is local maximum/minimum if the value of the determinant is positive. Otherwise, the critical point is the saddle point. In this problem, the Hessian matrix is

⌈-2y 3-2x-2y⌉

⌊3-2x-2y -2x⌋.

At point (0,0), we have

⌈-2y 3-2x-2y⌉ = ⌈0 3⌉

⌊3-2x-2y -2x⌋ ⌊3 0⌋.

The determinant is -9<0. So (0,0) is a saddle point.

At point (1,1), we have

⌈-2y 3-2x-2y⌉ = ⌈-2 -1⌉

⌊3-2x-2y -2x⌋ ⌊-1 -2⌋.

The determinant is 3>0. So (1,1) is a local maximum/minimum. With the fact that ∂²f/∂x²=-2<0, we know (1,1) is a local maximum, with value f(1,1)=1. (This would be local minimum if ∂²f/∂x²>0).

At point (3,0) and point (0,3), both determinants are negative. So they are saddle points.

Once we sort out the local maximum/minimum, we need to consider the boundary.

On the boundary, the function is reduced to one variable function. We have three lines on the boundary, x = 0, y = 0, and x+y = 4.

Along x = 0, we have f(0,y) = 0, a constant function. Along y = 0, we have f(x,0) = 0.

Along x+y = 4, we have f(x, 4-x) = -x(4-x), a function on variable x. To clarify things, let us call this function g(x) = -x(4-x), a quadratic function. Again, we need to find the local maximum/minimum for g(x). The function g(x) has a minimum at x=2, with value -4. The hidden condition here is that 0≤x≤4, because of the boundary we have for f(x). In other words, g(x) has two local maximums at x=0 and x=4, with g(0)=g(4)=0. Returning back to f(x), the above computation tells us that f(x) along x+y=4 has local minimum at (2,2) with f(2,2)=-4, and local maximums at (0,4), (4,0) with f(0,4)=f(4,0)=0.

Summerize everything. We have local maximums at (1,1) with f(1,1)=1. Along the boundary, we have local maximums at {(x,0),(0,y), 0≤x,y≤4} and we have local minimums at (2,2) with f(2,2)=-4. Compare them, we can say, the abesolute maximum of f(x,y) in R is (1,1) and the abesolute minimum of f(x,y) in R is (2,2).

Answer 2

We can do this in the traditional way, but why not use a little logic.

For x + y = c, c >= 0, xy is maximized with x = y = c/2

xy >= 0 for x, y >= 0.

We can consider this problem to be to find the maximum and minimum of

x^2(3 - 2x) for 0 <= x <= 2

f(x) = 3x^2 - 2x^3

f'(x) = 6x - 6x^2 = 6x(1 - x)

Zeroes are at 0 and 1. We only need to consider the 0 at x = 1

f''(x) = 6 - 12x

f''(0) = -6

Thus, f(x) is a maximum at 1.

1^2(3-2(1)) = 1*1 = 1

f(0) = 0(3) = 0

f(2) = 2^2(3-2*2) = 4*-1 = -4

f(x, y) i a maximum of 1 at x = 1, y = 1

f(x, y) is a minimum of -4 at x = 2, y = 2