For each of the following functions, find the maximum and minimum values of the function on the rectanglar region: −1≤x≤1,−2≤y≤2. Do this by looking at level curves and gradients. (A) f(x, y)=x+y+3: maximum value = minimum value = (B) f(x, y)=3x2+4y2: maximum value = minimum value = (C) f(x, y)=4x2−y2: maximum value = minimum value =

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)

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step-by-step explanation: