The functIonu(x,y) =x2+xy+ 3x+ 2y+ 5has a point of :a)both max and min...
Let z = x2 + xy + 3x + 2y + 5
For max. or min of z, we must have
at (–2, 1)
rt – s2 = –1 < 0
∴ neither max. nor min.
The correct answer is: neither max nor min
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The functIonu(x,y) =x2+xy+ 3x+ 2y+ 5has a point of :a)both max and min...
Explanation:
To determine whether the given function has a maximum or minimum point, we can use the method of partial derivatives. By finding the partial derivative of the function with respect to each variable, we can determine the critical points of the function.
Step 1: Find the partial derivative with respect to x:
To find the partial derivative of the function with respect to x, we differentiate each term in the function with respect to x while treating y as a constant.
Taking the partial derivative of each term:
∂u/∂x = 2x + y + 3
Step 2: Find the partial derivative with respect to y:
To find the partial derivative of the function with respect to y, we differentiate each term in the function with respect to y while treating x as a constant.
Taking the partial derivative of each term:
∂u/∂y = x + 2
Step 3: Set the partial derivatives equal to zero:
To find the critical points, we set the partial derivatives equal to zero and solve for x and y.
Setting ∂u/∂x = 0:
2x + y + 3 = 0
Setting ∂u/∂y = 0:
x + 2 = 0
Solving these equations simultaneously, we find the values of x and y:
x = -2
y = 1
Step 4: Determine whether the critical point is a maximum, minimum, or neither:
To determine whether the critical point is a maximum, minimum, or neither, we can use the second partial derivative test. This involves finding the second partial derivatives of the function and evaluating them at the critical point.
Taking the second partial derivative with respect to x:
∂²u/∂x² = 2
Taking the second partial derivative with respect to y:
∂²u/∂y² = 0
Taking the mixed partial derivative:
∂²u/∂x∂y = 1
Evaluating the second partial derivatives at the critical point (x = -2, y = 1):
∂²u/∂x² = 2
∂²u/∂y² = 0
∂²u/∂x∂y = 1
The second partial derivative test states that if the second partial derivatives ∂²u/∂x² and ∂²u/∂y² have different signs at the critical point, then the critical point is a saddle point and not a maximum or minimum. In this case, since ∂²u/∂x² = 2 > 0 and ∂²u/∂y² = 0, the signs are different and the critical point is a saddle point.
Therefore, the function does not have a maximum or minimum point. Hence, the correct answer is option D: neither max nor min.