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Chapter 7: Problem 18

Find all points \((x, y)\) where \(f(x, y)\) has a possible relative maximum or minimum. Then, use the second-derivative test to determine, if possible, the nature of \(f(x, y)\) at each of these points. If the second-derivative test is inconclusive, so state. $$ f(x, y)=2 x^{2}+3 x y+5 y^{2} $$

Short Answer

Expert verified
The function has a relative minimum at (0, 0).

Step by step solution

01

- Find critical points

To find the critical points, calculate the partial derivatives of the function and set them equal to zero. \(f_x = \frac{\partial f}{\partial x} = 4x + 3y\)\(f_y = \frac{\partial f}{\partial y} = 3x + 10y\). Setting these equal to zero: \(4x + 3y = 0\)\(3x + 10y = 0\).
02

- Solve for critical points

Solve the system of equations: \[4x + 3y = 0\]\[3x + 10y = 0\]. From the first equation, isolate y: \(y = -\frac{4}{3}x\). Substitute this into the second equation: \(3x + 10(-\frac{4}{3}x) = 0\) \(3x - \frac{40}{3}x = 0\) \(-\frac{31}{3}x = 0\)\(x = 0\) Substitute x back to find y: \(y = -\frac{4}{3}(0) = 0\). The critical point is at \((0, 0)\).
03

- Calculate second-order partial derivatives

Calculate the second partial derivatives: \(f_{xx} = \frac{\partial^2 f}{\partial x^2} = 4\)\(f_{yy} = \frac{\partial^2 f}{\partial y^2} = 10\)\(f_{xy} = \frac{\partial^2 f}{\partial x \partial y} = 3\).
04

- Determine nature using the second-derivative test

Use the second-derivative test: The discriminant is given by: \(D = f_{xx} f_{yy} - (f_{xy})^2\).Substitute the values: \(D = 4 \cdot 10 - 3^2 = 40 - 9 = 31\).Since \(D > 0\) and \(f_{xx} > 0\), there is a relative minimum at \((0, 0)\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Critical Points
Partial Derivatives are used to understand how a function changes as each variable changes, holding other variables constant. These derivatives are crucial when dealing with functions of several variables. In the context of our function f(x, y)=2 x^{2}+3 x y+5 y^{2}, the partial derivatives are: f_x = 4x + 3y and f_y = 3x + 10y. When we set these partial derivatives equal to zero to find the critical points, it helps us identify where the function may reach local maxima, minima, or saddle points. Partial derivatives also enable us to form second partial derivatives, which are necessary for the second-derivative test.
Relative Extrema
Relative extrema refer to points in the domain of a function where the function reaches a local maximum or minimum. After finding the critical points using the first partial derivatives, to determine whether these points are maxima, minima, or saddle points, we employ the second-derivative test. For our function f(x, y)=2 x^{2}+3 x y+5 y^{2}, we calculate the second-order partial derivatives: f_{xx} = 4, f_{yy} = 10, and f_{xy} = 3. The discriminant, D = f_{xx} f_{yy} - (f_{xy})^2, allows us to categorize the critical point at (0, 0). Since D = 31 > 0 and f_{xx} > 0, it indicates that (0, 0) is a relative minimum for our function.

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Most popular questions from this chapter

A company manufactures and sells two products, I and II, that sell for \(\$ p_{\mathrm{I}}\) and \(\$ p_{\mathrm{II}}\) per unit, respectively. Let \(C(x, y)\) be the cost of producing \(x\) units of product \(\mathrm{I}\) and \(y\) units of product II. Show that if the company's profit is maximized when \(x=a, y=b\), then $$ \frac{\partial C}{\partial x}(a, b)=p_{\mathrm{I}} \quad \text { and } \quad \frac{\partial C}{\partial y}(a, b)=p_{\mathrm{II}} $$.

A farmer can produce \(f(x, y)=200 \sqrt{6 x^{2}+y^{2}}\) units of produce by utilizing \(x\) units of labor and \(y\) units of capital. (The capital is used to rent or purchase land. materials, and equipment.) (a) Calculate the marginal productivities of labor and capital when \(x=10\) and \(y=5\). (b) Let \(h\) be a small number. Use the result of part (a) to determine the approximate effect on production of changing labor from 10 to \(10+h\) units while keeping capital fixed at 5 units. (c) Use part (b) to estimate the change in production when labor decreases from 10 to \(9.5\) units and capital stays fixed at 5 units.

A company manufactures and sells two products, I and II, that sell for \(\$ 10\) and \(\$ 9\) per unit, respectively. The cost of producing \(x\) units of product I and \(y\) units of product II is $$ 400+2 x+3 y+.01\left(3 x^{2}+x y+3 y^{2}\right) . $$ Find the values of \(x\) and \(y\) that maximize the company's profits. [Note: Profit \(=(\) revenue \()-(\) cost \() .\).

Let \(f(x, y)=3 x^{2}+2 x y+5 y\), as in Example \(5 .\) Show that $$ f(1+h, 4)-f(1,4)=14 h+3 h^{2} . $$ Thus, the error in approximating \(f(1+h, 4)-f(1,4)\) by \(14 h\) is \(3 h^{2}\). (If \(h=.01\), for instance, the error is only \(.0003 .\)

Solve the following exercise by the method of Lagrange multipliers. Minimize \(x^{2}+3 y^{2}+10\), subject to the constraint \(8-x-y=0 .\)
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