Find the absolute maximum and minimum of the function f(x,y)=ysqrt(x)−y^2−x+3y on the domain 0≤x≤9, 0≤y≤6.
Fx = y/(2√x) - 1
Fy = √x - 2y+3
Fx = Fy = 0 at (1,2)
But this is a maximum.
So, to find the minimum, check f(x,y) at the four corners of the domain.
f(0,6) = -18
To find the absolute maximum and minimum of the function f(x, y) = y√(x) − y^2 − x + 3y on the domain 0 ≤ x ≤ 9, 0 ≤ y ≤ 6, we need to follow these steps:
1. Find the critical points by taking the partial derivatives of f(x, y) with respect to x and y, and set them equal to zero.
∂f/∂x = (y/2√(x)) - 1 = 0
∂f/∂y = √(x) - 2y + 3 = 0
2. Solve the above system of equations to find the critical points.
From equation 1, we get y/2√(x) = 1, which implies y = 2√(x).
Substituting this value of y into equation 2, we get √(x) - 4√(x) + 3 = 0.
Solving this quadratic equation, we find x = 1 and x = 9.
Substituting these values of x back into equation 1, we can find the corresponding values of y: y = 2 for x = 1, and y = 6 for x = 9.
3. Evaluate the function f(x, y) at the critical points and at the endpoints of the domain.
The function values at the critical points are:
f(1, 2) = 2√(1) - 2^2 - 1 + 3(2) = 2 - 4 - 1 + 6 = 3,
f(9, 6) = 6√(9) - 6^2 - 9 + 3(6) = 54 - 36 - 9 + 18 = 27.
The function values at the endpoints of the domain are:
f(0, 0) = 0 - 0 - 0 + 3(0) = 0,
f(0, 6) = 6√(0) - 6^2 - 0 + 3(6) = 0 - 36 + 18 = -18,
f(9, 0) = 0 - 0 - 9 + 3(0) = -9,
f(9, 6) = 6√(9) - 6^2 - 9 + 3(6) = 54 - 36 - 9 + 18 = 27.
4. Comparing the above function values, we can determine the absolute maximum and minimum.
The absolute maximum value is 27, which occurs at (9, 6).
The absolute minimum value is -18, which occurs at (0, 6).
Therefore, the absolute maximum and minimum of the function f(x, y) = y√(x) - y^2 - x + 3y on the domain 0 ≤ x ≤ 9, 0 ≤ y ≤ 6 are:
Absolute maximum: f(9, 6) = 27
Absolute minimum: f(0, 6) = -18
To find the absolute maximum and minimum of the function f(x, y) = y√(x) - y^2 - x + 3y on the given domain 0 ≤ x ≤ 9, 0 ≤ y ≤ 6, we can follow these steps:
Step 1: Calculate the critical points of the function
- To find the critical points, we need to take the partial derivatives of f with respect to x and y and set them equal to zero.
Let's start by finding the partial derivative with respect to x:
∂f/∂x = -1/2y√(x) - 1
And now, let's find the partial derivative with respect to y:
∂f/∂y = √(x) - 2y + 3
- Setting ∂f/∂x = 0, we have:
-1/2y√(x) - 1 = 0
-1/2y√(x) = 1
y√(x) = -2
From this equation, we can see that there are no real solutions where y is greater than zero, since the square root of x is always non-negative.
- Setting ∂f/∂y = 0, we have:
√(x) - 2y + 3 = 0
2y = √(x) + 3
y = (1/2)√(x) + 3/2
From this equation, we can see that the value of y increases as x increases, which indicates that y is highest when x is highest. Since the domain is bounded by 0 ≤ x ≤ 9, 0 ≤ y ≤ 6, the maximum value of y is 6.
Therefore, the critical point is at (9, 6).
Step 2: Evaluate the function at the boundary points and the critical point
- Evaluate the function at the boundary points of the domain:
- At x = 0:
f(0, y) = y√(0) - y^2 - 0 + 3y
f(0, y) = 0 - y^2 + 3y
- At x = 9:
f(9, y) = y√(9) - y^2 - 9 + 3y
f(9, y) = 3y - y^2 - 9 + 3y
- Evaluate the function at the critical point (9, 6):
f(9, 6) = 6√9 - 6^2 - 9 + 3(6)
f(9, 6) = 18 - 36 - 9 + 18
f(9, 6) = -9
Step 3: Compare the values obtained in step 2 to find the absolute maximum and minimum
- Comparing the values obtained:
- At x = 0: f(0, y) = -y^2 + 3y
- At x = 9: f(9, y) = 3y - y^2 - 9 + 3y
- At (9, 6): f(9, 6) = -9
- The maximum value is obtained at (9, 6): f(9, 6) = -9
- The minimum value can be obtained by comparing the values at x = 0 and x = 9.
- At x = 0: f(0, y) = -y^2 + 3y
- At x = 9: f(9, y) = 3y - y^2 - 9 + 3y
By comparing these values, we can determine the minimum.
Therefore, the absolute maximum value of f(x, y) = y√(x) - y^2 - x + 3y on the given domain is -9, and the absolute minimum value can be determined by comparing the values obtained at the boundary points.