Find the maximum value of the function f(x,y) = 9x2−9xy+y3 on the closed rectangle whose corners are the points (1,1), (1,−1), (−1,1) and (−1,−1)
f(-1,y) = y^3+9y+9
f(1,y) = y^3-9y+9
f(x,-1) = 9x^2+9x-1
f(x,1) = 9x^2-9x+1
So, check for maxima of those on the interval [-1,1]
Then, you have
∂f/∂x = 18x-9y
∂f/∂y = 3y^2-9x
Extrema occur when both these are zero.
Along the line ∂f/∂x=0 or y=2x,
g(x) = f(x,y) = 8x^3-9x^2
g'(x) = 24x^2-18x = 6x(4x-3)
So, at (3/4,3/2) there is an extremum
You can do the other case, and use the 2nd derivative tests to determine whether it's a max or min.
To find the maximum value of the function f(x, y) = 9x^2 - 9xy + y^3 on the closed rectangle, we need to evaluate the function at all the critical points inside the rectangle and compare the values.
To begin, let's find the critical points of the function. Critical points occur when the partial derivatives of the function are equal to zero.
Step 1: Find the partial derivative with respect to x:
f_x = d/dx(9x^2 - 9xy + y^3)
= 18x - 9y
Step 2: Find the partial derivative with respect to y:
f_y = d/dy(9x^2 - 9xy + y^3)
= -9x + 3y^2
Setting both partial derivatives equal to zero and solving the system of equations, we get:
18x - 9y = 0 -- (1)
-9x + 3y^2 = 0 -- (2)
From equation (1), we can solve for y in terms of x:
y = 2x
Substituting this into equation (2), we have:
-9x + 3(2x)^2 = 0
-9x + 12x^2 = 0
3x(4x - 3) = 0
So, we have two possible critical points:
1. x = 0, y = 0
2. x = 3/4, y = 3/2
Now, let's evaluate the function f(x, y) at the corners of the rectangle and the critical points to compare the values:
Corner 1: (1, 1)
f(1, 1) = 9(1)^2 - 9(1)(1) + (1)^3 = 9 - 9 + 1 = 1
Corner 2: (1, -1)
f(1, -1) = 9(1)^2 - 9(1)(-1) + (-1)^3 = 9 + 9 - 1 = 17
Corner 3: (-1, 1)
f(-1, 1) = 9(-1)^2 - 9(-1)(1) + (1)^3 = 9 + 9 + 1 = 19
Corner 4: (-1, -1)
f(-1, -1) = 9(-1)^2 - 9(-1)(-1) + (-1)^3 = 9 - 9 - 1 = -1
Critical Point 1: (0, 0)
f(0, 0) = 9(0)^2 - 9(0)(0) + (0)^3 = 0
Critical Point 2: (3/4, 3/2)
f(3/4, 3/2) = 9(3/4)^2 - 9(3/4)(3/2) + (3/2)^3
Now, we compare the values:
f(1, 1) = 1
f(1, -1) = 17
f(-1, 1) = 19
f(-1, -1) = -1
f(0, 0) = 0
f(3/4, 3/2) = ?
By evaluating f(3/4, 3/2), we can determine if it is greater than the other values we have calculated.