1. Find the maximum of f(x,y) = x² + xy + 2y² - 3x + 2y on the triangle

with vertices (0,0), (0,1) and (1,0).

First, find the min/max points of the function:

F_x = 2x + y – 3

F_y = 4y + x + 2

Where F_x and F_y are the partial derivatives of f(x,y) with respect to x,y respectively. Setting both to zero yields:

y = 3 – 2x

12 – 8x + x +2 = 0

14 – 7x = 0

x = 2 , y = -1

This point does not lie in the triangle, so we need to look for local min/max along the edges of the triangle, that is the lines (0,0) -> (1,0), (0,0) -> (0,1), and (1,0) -> (0,1):

On the (0,0) -> (1,0) line, y=0 everywhere, so f(x,y=0) = g(x) = x² – 3x

g’(x) = 2x – 3 so x = 3/2 which lies outside the (0,1) interval, so f(x,y) has a local min/max at the endpoints of the interval, and we will check those later.

On the (0,0) -> (0,1) line, x=0 everywhere, so f(x=0,y) = h(y) = 2y² + 2

h’(y) = 4y + 2 so y = -1/2 which lies outside the (0,1) interval as well.

On the (1,0) ->(0,1) line, y = 1 – x, so let

f(x,y = 1-x) = k(x) = x² + x(1-x) +2 (1-x)² -3x + 2(1-x)

k(x) = x² + x – x² + 2(1 – 2x –x²) - 3x +2 -2x = -4x +2 +2 -4x +2x²

k(x) = 2x² - 8x + 2

k’(x) = 4x – 8 so x = 2 which lies outside the (0,1) interval as well.

So all critical points of the function f(x,y) = x² + xy + 2y² - 3x + 2y have to be at the corners, so just check the values of the function there:

f(0,0) = 0

f(1,0) = 1 – 3 = -2

f(0,1) = 2 + 2 = 4

so the function has a maximum at the (0,1) corner (and a minimum a the (1,0) corner).

2. Maximize f(x, y) = 3xy + 1 on the ellipse (x/a)² + (y/b)² = 1 (with a, b >0).

Although we could substitute in for x or y here as well, it is much better to use the method of Lagrange multipliers:

L = 3xy + 1 – λ ((x/a)² + (y/b)² - 1)

L_x = 3y – 2 λ x / a²

L_y = 3x – 2 λ y / b²

L_λ = (x/a)² + (y/b)² – 1

Where L_x, L_y, L_λ are the partial derivatives of L wtr. x,y, and λ respectively.

Setting all three to equal zero and solving L_x for λ we get

plugging into L_y we get

which yields

plugging into L_λ we get

which yields

or and thus solving for x we get

So the function has four critical points on the ellipse, and we can easily see the maxima are at () and ().