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How to Find the Absolute Maximum and the Absolute Minimum Values of a Continuous Function on a Closed and Bounded Region in the Plane
Theorem: Let f (x, y) be a continuous function defined on a closed and bounded region R in the plane. Then f has an absolute maximum and an absolute minimum value on R.
This theorem is the analogue of the following theorem for 1-variable functions:
Theorem: Let f (x) be a continuous function defined on a closed interval of finite length [a, b]. Then f has an absolute maximum and an absolute minimum value on [a, b].
Let us recall how we find these values in the 1-variable case:
How to find the absolute maximum and the absolute minimum values of a continuous function f (x) on a closed interval of finite length [a, b]:
- Compute f โฒ.
- Find the critical points of f in (a, b).
- Add the endpoints a and b of the interval [a, b] to the list of points found in step 2.
- Compute the value of f at each of the points in the list.
- The largest value in step 4 is the absolute maximum value of f on [a, b] and the smallest value is the absolute minimum value.
There is a similar algorithm in the 2-variable case:
How to find the absolute maximum and the absolute minimum values of a continuous function f (x, y) on a closed and bounded region R:
- Compute fx and fy.
- Find the critical points of f in the interior of R.
- Find the critical points of the restriction of f (x, y) to the boundary of R and add these to the list of points found in step 2.
- Compute the value of f at each of the points in the list.
- The largest value in step 4 is the absolute maximum value of f on R and the smallest value is the absolute minimum value.
Example: Find the the absolute maximum and the absolute minimum values of f (x, y) = x^3 โ xy + y^2 โ x on R = {(x, y) : x โฅ 0 , y โฅ 0 and x + y โค 2 }.
We apply the algorithm:
- fx = 3x^2 โ y โ 1 and fy = โx + 2y.
- We have to solve the equations 3x^2 โ y โ 1 = 0 and โx + 2y = 0 simultaneously. From the second one we solve for x to get x = 2y and substitute it in the first one to get 3 ยท (2y)^2 โ y โ 1 = 0 =โ 12 y^2 โ y โ 1 = 0 =โ y = 1/3 or y = โ 1 /4. Since x = 2y, these give the x-values y = 1/3 =โ x = 2/3 and y = โ 1 /4 =โ x = โ 1 /2. So we obtain the points (x, y) = (2/ 3 , 1 /3) and (x, y) = (โ 1 / 2 , โ 1 /4). Note that the second point is not in R. So we put only the first one, (2/ 3 , 1 /3) in our list.
This is important! If you include points that are not in R in your list, there is no guarantee that the algorithm will give the correct answer.
- Now we come to the boundary. The boundary of R is the union of three sides of the triangle: Side I = {(x, y) : 0 โค x โค 2 and y = 0}, Side II = {(x, y) : x = 0 and 0 โค y โค 2 } and Side III = {(x, y) : x + y = 2 and 0 โค x โค 2 }. So we have essentially three 1-variable problems to solve:
Side I : We parametrize this side in the obvious way: x = t, y = 0 for 0 โค t โค 2. Then the restrictions of f to Side I is f (t, 0) = t^3 ยท 0 โ t ยท 0 โ 02 โ t = t^3 โ t for 0 โค t โค 2. Now we find the list of points for this 1-variable optimization problem: d/dt(f (t, 0)) = 3t^2 โ 1 = 0 =โ t = 1/
3 or t = โ 1
- The second solution is not in the interval [0, 2], so we only take the first one t = 1/
3, and the endpoints of the interval [0, 2], t = 0 and t = 2. These correspond to the points (x, y) = (1/
(x, y) = (0, 0), and (x, y) = (2, 0).
Side II : This is similar to the previous calculation: f (0, t) = t^2 for 0 โค t โค 2. d/dt(f (0, t)) = 2t = 0 =โ t = 0. This is one of the endpoints, the other one is t = 2, and they correspond to the points (x, y) = (0, 0) and (x, y) = (0, 2) in the plane.
Side III : This time one possible parametrization is x = t, y = 2 โ t for 0 โค t โค 2. This gives f (t, 2 โ t) = t^3 + 2t^2 โ 7 t + 4 for 0 โค t โค 2. Then d/dt(f (t, 2 โ t)) = 3t^2 + 4 t โ 7 = 0 =โ t = 1 or t = โ 7 /3. Again the second solution is not in the interval we are looking at. The first one gives t = 1 =โ (x, y) = (t, 2 โ t) = (1, 1). So our points, including the endpoints, are (x, y) = (1, 1), (x, y) = (2, 0) and (x, y) = (0, 2).
To summarize, we found the points
(0, 0), (2, 0), (0, 2), (1/
in this step. So now the complete list of points we are going to look at is
(2/ 3 , 1 /3), (1/
