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Math  /  Calculus

Question1. Исследовать функции на локальный экстремум. Указать точки экстремума, вид экстремума, значение функции в точках экстремума. (по 1 баллу)
Ответы: 1.1z=3x+6y−x2−xy+y21.1 z=3 x+6 y-x^{2}-x y+y^{2} экстремумов нет

Studdy Solution

STEP 1

1. We are given a function z=3x+6y−x2−xy+y2 z = 3x + 6y - x^2 - xy + y^2 .
2. We need to find local extrema, if any, and specify the type of extremum, the points of extremum, and the function values at those points.

STEP 2

1. Find the critical points of the function.
2. Use the second derivative test to determine the nature of the critical points.
3. Conclude whether there are any local extrema and specify their details.

STEP 3

To find the critical points, calculate the first partial derivatives of z z with respect to x x and y y .
∂z∂x=3−2x−y\frac{\partial z}{\partial x} = 3 - 2x - y
∂z∂y=6−x+2y\frac{\partial z}{\partial y} = 6 - x + 2y
Set the partial derivatives equal to zero to find critical points:
3−2x−y=03 - 2x - y = 0
6−x+2y=06 - x + 2y = 0

STEP 4

Solve the system of equations:
From 3−2x−y=0 3 - 2x - y = 0 , we get y=3−2x y = 3 - 2x .
Substitute y=3−2x y = 3 - 2x into 6−x+2y=0 6 - x + 2y = 0 :
6−x+2(3−2x)=06 - x + 2(3 - 2x) = 0
6−x+6−4x=06 - x + 6 - 4x = 0
12−5x=012 - 5x = 0
x=125x = \frac{12}{5}
Substitute x=125 x = \frac{12}{5} back into y=3−2x y = 3 - 2x :
y=3−2(125)=3−245=155−245=−95y = 3 - 2\left(\frac{12}{5}\right) = 3 - \frac{24}{5} = \frac{15}{5} - \frac{24}{5} = -\frac{9}{5}
The critical point is (125,−95) \left( \frac{12}{5}, -\frac{9}{5} \right) .

STEP 5

To determine the nature of the critical point, use the second derivative test. Compute the second partial derivatives:
∂2z∂x2=−2\frac{\partial^2 z}{\partial x^2} = -2
∂2z∂y2=2\frac{\partial^2 z}{\partial y^2} = 2
∂2z∂x∂y=−1\frac{\partial^2 z}{\partial x \partial y} = -1
Calculate the determinant of the Hessian matrix:
D=(∂2z∂x2)(∂2z∂y2)−(∂2z∂x∂y)2D = \left( \frac{\partial^2 z}{\partial x^2} \right) \left( \frac{\partial^2 z}{\partial y^2} \right) - \left( \frac{\partial^2 z}{\partial x \partial y} \right)^2
D=(−2)(2)−(−1)2=−4−1=−5D = (-2)(2) - (-1)^2 = -4 - 1 = -5
Since D<0 D < 0 , the critical point (125,−95) \left( \frac{12}{5}, -\frac{9}{5} \right) is a saddle point, not a local extremum.

STEP 6

Since the determinant of the Hessian is negative, there are no local extrema for the given function.
Conclusion: The function has no local extrema.

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