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Problem 152

Find the critical points and the nature of each critical point (i.e., relative maximum, relative minimum, or saddle point) for: a) $$f(x, y)=x^{2}-2 x y+2 y^{2}+x-5$$ b) $\mathrm{f}(\mathrm{x}, \mathrm{y})=(1-\mathrm{x})(1-\mathrm{y})(\mathrm{x}+\mathrm{y}-1)$.

Expert verified
The critical points and their nature for the given functions are: a) For $$f(x, y)=x^{2}-2 x y+2 y^{2}+x-5$$, there is one critical point at $$\left(-\frac{1}{2}, 0\right)$$, which is a relative minimum. b) For $$f(x, y)=(1-x)(1-y)(x+y-1)$$, there are two critical points at $$(0,0)$$ and $$(1,1)$$. However, the second partial derivatives test is inconclusive, and we cannot determine their nature using this method.
See the step by step solution

Step 1: Find first-order partial derivatives

Compute the partial derivatives of the function with respect to x and y. $f_x = \frac{\partial f}{\partial x} = 2x - 2y + 1$ $f_y = \frac{\partial f}{\partial y} = -2x + 4y - 1$

Step 2: Set partial derivatives equal to zero

To find the critical points, we need to solve the following system of equations: $2x - 2y + 1= 0$ $-2x + 4y - 1= 0$

Step 3: Solve the system of equations

We can solve the system of equations by substitution or elimination method. Here, we use the elimination method. $2(2x - 2y + 1) + (-2x + 4y - 1) = 0$ $4x - 4y + 2 - 2x + 4y - 1 = 0$ $2x + 1 = 0$ $x = -\frac{1}{2}$ Now, substitute x back into one of the original equations: $-2\left(-\frac{1}{2}\right) + 4y - 1= 0$ $1 + 4y - 1= 0$ $4y = 0$ $y = 0$ The critical point is $$\left(-\frac{1}{2}, 0\right)$$.

Step 4: Analyze the critical point

To determine the nature of the critical point, we will compute the second-order partial derivatives and use the discriminant test. Second-order partial derivatives: $f_{xx} = \frac{\partial^2 f}{\partial x^2} = 2$ $f_{xy} = \frac{\partial^2 f}{\partial x\partial y} = -2$ $f_{yy} = \frac{\partial^2 f}{\partial y^2} = 4$ Discriminant: $D = f_{xx}f_{yy} - f_{xy}^2 = (2)(4) - (-2)^2 = 8 - 4 = 4$ Since $$D > 0$$ and $$f_{xx} > 0$$, the critical point $$\left(-\frac{1}{2}, 0\right)$$ is a relative minimum. b) $$f(x, y)=(1-x)(1-y)(x+y-1)$$

Step 1: Find first-order partial derivatives

Compute the partial derivatives of the function with respect to x and y. $f_x = (1 - y)(x + y - 1) - (1 - x)(1 - y)$ $f_y = (1 - x)(x + y - 1) - (1 - x)(1 - y)$

Step 2: Set partial derivatives equal to zero

To find the critical points, we need to solve the following system of equations: $(1 - y)(x + y - 1) - (1 - x)(1 - y)= 0$ $(1 - x)(x + y - 1) - (1 - x)(1 - y)= 0$

Step 3: Solve the system of equations

We can solve the system of equations by expressing x and y in terms of a single variable. Since both equations look almost the same, we can use the similarity to equate them and solve for the unknown variable. Here, we equate both equations: $(1 - y)(x + y - 1) - (1 - x)(1 - y)= (1 - x)(x + y - 1) - (1 - x)(1 - y)$ Simplifying the equation, we get: $x = y$ Now, substitute x = y back into one of the original equations: $(1 - y)(y + y - 1) - (1 - y)(1 - y) = 0$ $2y(1 - y) - (1 - y)^2 = 0$ Solve for y: $y(2 - 2y - 1 + 2y) = (1 - y)(1 - y)= 0$ $y = 0 \quad \text{or} \quad y = 1$ The critical points are $$(0,0)$$ and $$(1,1)$$.

Step 4: Analyze the critical points

To determine the nature of the critical points, we will compute the second-order partial derivatives and use the discriminant test. First, simplify the expressions for the first-order derivatives: $f_x = x - y$ $f_y = y - x$ Second-order partial derivatives: $f_{xx} = \frac{\partial^2 f}{\partial x^2} = 1$ $f_{xy} = \frac{\partial^2 f}{\partial x\partial y} = -1$ $f_{yy} = \frac{\partial^2 f}{\partial y^2} = 1$ Discriminant: $D = f_{xx}f_{yy} - f_{xy}^2 = (1)(1) - (-1)^2 = 1 - 1 = 0$ Since $$D = 0$$, the discriminant test is inconclusive. We cannot determine the nature of the critical points $$(0,0)$$ and $$(1,1)$$ using the second partial derivatives.

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