Answer:
A) The maximum point is 6 at (1,2) while minimum is 0 at (-1,-2)
B) The maximum is 19 at (-1,2), (1,-2),(1,2) & minimum is 0 at (0 , 0)
C) The maximum and minimum is 0 at (+/- 1, +/- 2) and (0,0)
Step-by-step explanation:
From the question, the region is;
−1≤x≤1, −2≤y≤2
In order for us to find the maximum or minimum of a function, we need to find the critical points by finding the first partial derivative and then second derivative if needed;
Thus;
A) f(x,y) = x+y+3
f'(x) = 1 and f'(y)= 1
Therefore, we have no critical points and so we are going to check only the endpoints into the original equation as the follows;
x+y+3;
f(-1 , -2) = -1 + (-2) + 3 = -3 + 3 = 0
f(-1 , 2) = -1 + 2 + 3 = 3 + 1 = 4
f(1 , -2) = 1 + (-2) + 3 = -1 + 3 = 2
f(1 , 2) = 1 + 2 + 3 = 3 + 3 = 6
From the values gotten, we can see that, the maximum value is 6 while the minimum is 0. Thus the maximum point is (1,2) while minimum is at (-1,-2)
B) f(x,y) = 3x² + 4y²
f'(x) = 6x and f'(y)= 8y
If we equate each of them to zero to find the critical point, we'll obtain:
6x = 0 and so, x = 0
8y = 0 and so, y = 0
So, critical point is at (0 , 0)
Now, let's find the second derivative to check ;
f''x = 6, f''xy = 0, f''yy = 8
D = (f''x)(f''y) - (f''xy)² = (6 x 8) - (0)² = 48
Since D(0, 0) = 48 > 0 and f''x (0, 0) = 0, therefore it might be maximum or minimum and we can't say which it is.
So, let's check the critical point with the endpoints into the original equation to obtain:
f(0 , 0) = 3(0²) + 4(0²) = 0
f(-1 , -2) = 3(-1²) + 4(-2)² = 3 + 16 = 19
f(-1 , 2) = 3(-1²) + 4(2)² = 3 + 16 = 19
f(1 , -2) = 3(1²) + 4(-2)² = 3 + 16 = 19
f(1 , 2) = 3(1²) + 4(2)² = 3 + 16 = 19
Thus maximum occurs at (-1,2), (1,-2),(1,2) & minimum occurs at (0 , 0)
C)f(x , y) = 4x² - y²
Repeating the same process as B above;
f'x = 8x
f'y = -2y
If we equate each of them to zero to find the critical point, we'll obtain:
8x = 0 and so, x = 0
-2y = 0 and so, y = 0
So, critical point is at (0 , 0)
Now, let's find the second derivative to check ;
f''x = 8, f''xy = 0, f''y = -2
D = (f''x)(f''y) - (f''xy)² = (8 x 2) - (0)²= 16
Since D(0, 0) = 16 > 0 and f''x (0, 0) = 0, therefore it might be maximum or minimum and we can't say which it is.
So, let's check the critical point with the endpoints into the original equation to obtain:
f(0 , 0) = 4(0)² - 0² = 0
f(-1 , -2) = 4(-1)² - (-2)² = 4 - 4 = 0
f(-1 , 2) = 4(-1)² - 2² = 4 - 4 = 0
f(1 , -2) = 4(1)² - (-2)² = 4 - 4 = 0
f(1 , 2) = 4(1)² - (2)² = 4 - 4 = 0
Looking at the values, maximum and minimum is 0 at (+/- 1, +/- 2) and (0,0)
