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Use Lagrange multipliers to find these values. $$f(x, y)=x^{2} y \text { subject to } x^{2}+y^{2}=9$$

Short Answer

Expert verified
Based on the problem and the step-by-step solution, the answer to the question is: The largest value of the function f(x, y) = x^2y subject to the constraint x^2 + y^2 = 9 is $$\frac{9\sqrt{2}}{2}$$, and the smallest value is $$-\frac{9\sqrt{2}}{2}$$.

Step by step solution

01

Set up the Lagrange equation

Write the given function f(x, y) and the constraint g(x, y), and set up the Lagrange equation: $$L(x, y, \lambda) = f(x, y) - \lambda(g(x, y) - c)$$ In this case, we are given: $$f(x, y) = x^2y$$ $$g(x, y) = x^2 + y^2 = 9$$ So, the Lagrange equation is: $$L(x, y, \lambda) = x^2y - \lambda(x^2 + y^2 - 9)$$
02

Find the partial derivatives

We need to compute the partial derivatives of L with respect to x, y, and λ: $$\frac{\partial L}{\partial x} = 2xy - 2x\lambda$$ $$\frac{\partial L}{\partial y} = x^2 - 2y\lambda$$ $$\frac{\partial L}{\partial \lambda} = x^2 + y^2 - 9$$
03

Solve the partial derivatives for the stationary points

Now, equate each of the partial derivatives to zero and solve for the values of x, y, and λ: $$2xy - 2x\lambda = 0$$ $$x^2 - 2y\lambda = 0$$ $$x^2 + y^2 = 9$$ Among these equations, the third one is our constraint equation. Let's solve the first equation for λ: $$λ = \frac{xy}{x}$$ $$λ = y$$ Next, we substitute λ in the second equation: $$x^2 = 2y^2$$ $$2x^2 = x^2 + y^2 = 9$$ $$x^2 = \frac{9}{2}$$ $$y^2 = \frac{9}{2}$$ We can now find the values of x and y: $$x = \pm \frac{3}{\sqrt{2}}$$ $$y = \pm \frac{3}{\sqrt{2}}$$ There are four possible combinations of x and y: (3/√2, 3/√2), (3/√2, -3/√2), (-3/√2, 3/√2), and (-3/√2, -3/√2).
04

Compute the extreme values of the function

Now compute the function f(x, y) at each of these combinations and identify the largest and smallest values: $$f\left(\frac{3}{\sqrt{2}}, \frac{3}{\sqrt{2}}\right) = \frac{9\sqrt{2}}{2}$$ $$f\left(\frac{3}{\sqrt{2}}, -\frac{3}{\sqrt{2}}\right) = -\frac{9\sqrt{2}}{2}$$ $$f\left(-\frac{3}{\sqrt{2}}, \frac{3}{\sqrt{2}}\right) = -\frac{9\sqrt{2}}{2}$$ $$f\left(-\frac{3}{\sqrt{2}}, -\frac{3}{\sqrt{2}}\right) = \frac{9\sqrt{2}}{2}$$ The largest value of the function is $$\frac{9\sqrt{2}}{2}$$ and the smallest value is $$-\frac{9\sqrt{2}}{2}$$.

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Key Concepts

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

Constraint Optimization
Constraint optimization is a method used to find the maximum or minimum values of a function, given a specific constraint. One common technique employed for such problems is the method of Lagrange multipliers. In constraint optimization, the function to be optimized is known as the objective function, and the condition we adhere to is the constraint.
The technique of Lagrange multipliers aids in finding points where the gradient of the objective function is parallel to the gradient of the constraint. This happens when both gradients are multiplied by a common scalar, known as the Lagrange multiplier.
  • The objective function, in our exercise, is defined by \( f(x, y) = x^2y \).
  • The constraint is represented by \( g(x, y) = x^2 + y^2 - 9 = 0 \).
The relation between these gradients provides us the system of equations which are then solved to give the maximum or minimum values under the given constraint. By setting up and solving the Lagrange equation, we combine these factors to precisely locate the extreme values of interest, considering the constraint.
Extreme Values
Extreme values refer to the maximum and minimum values that a function may take, under certain circumstances. When optimizing functions using constraints, we're interested in evaluating these extreme points to find out how the function behaves.
The function in the exercise, \( f(x, y) = x^2y \), showcases different extreme values depending on the values of \(x\) and \(y\) that satisfy the constraint \(x^2 + y^2 = 9\).
  • The computation leads us to a set of potential solutions, or extreme points, notably \( \left(\frac{3}{\sqrt{2}}, \frac{3}{\sqrt{2}}\right) \) and its negatives or permutations.
Once these points are determined, the values of \(f(x, y)\) at each point are calculated, revealing the largest and smallest outputs possible given the constraint. Finding these values tells us how the function reacts to changes in variables while maintaining the constraint, thus illustrating practical behaviors of equations in real-world situations.
Partial Derivatives
Partial derivatives play a key role in multiple-variable calculus, especially in constraint optimization. A partial derivative demonstrates the direction and rate of change of a function as it varies with respect to one of its variables. In the realm of optimization, they help pinpoint where changes occur and how to adjust values to meet constraints.
For the function \(L(x, y, \lambda) = x^2y - \lambda(x^2 + y^2 - 9)\), computed partial derivatives with respect to \(x\), \(y\), and \(\lambda\) give us:
  • \(\frac{\partial L}{\partial x} = 2xy - 2x\lambda\)
  • \(\frac{\partial L}{\partial y} = x^2 - 2y\lambda\)
  • \(\frac{\partial L}{\partial \lambda} = x^2 + y^2 - 9\)
By solving these derivatives set to zero, we derive critical points of the function under the imposed constraint. The idea is to find where all of these equalities hold true, indicating potential extreme values. Partial derivatives thus help navigate the terrain of complex functions by allowing us to explore multi-faceted changes efficiently.

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Most popular questions from this chapter

Travel cost The cost of a trip that is \(L\) miles long, driving a car that gets \(m\) miles per gallon, with gas costs of \(\$ p /\) gal is \(C=L p / m\) dollars. Suppose you plan a trip of \(L=1500 \mathrm{mi}\) in a car that gets \(m=32 \mathrm{mi} / \mathrm{gal},\) with gas costs of \(p=\$ 3.80 / \mathrm{gal}\) a. Explain how the cost function is derived. b. Compute the partial derivatives \(C_{L}, C_{m^{\prime}}\) and \(C_{p^{\prime}}\). Explain the meaning of the signs of the derivatives in the context of this problem. c. Estimate the change in the total cost of the trip if \(L\) changes from \(L=1500\) to \(L=1520, m\) changes from \(m=32\) to \(m=31,\) and \(p\) changes from \(p=\$ 3.80\) to \(p=\$ 3.85\) d. Is the total cost of the trip (with \(L=1500 \mathrm{mi}, m=32 \mathrm{mi} / \mathrm{gal}\). and \(p=\$ 3.80\) ) more sensitive to a \(1 \%\) change in \(L,\) in \(m,\) or in \(p\) (assuming the other two variables are fixed)? Explain.

Line tangent to an intersection curve Consider the paraboloid \(z=x^{2}+3 y^{2}\) and the plane \(z=x+y+4,\) which intersects the paraboloid in a curve \(C\) at (2,1,7) (see figure). Find the equation of the line tangent to \(C\) at the point \((2,1,7) .\) Proceed as follows. a. Find a vector normal to the plane at (2,1,7) b. Find a vector normal to the plane tangent to the paraboloid at (2,1,7) c. Argue that the line tangent to \(C\) at (2,1,7) is orthogonal to both normal vectors found in parts (a) and (b). Use this fact to find a direction vector for the tangent line.

Prove that the level curves of the plane \(a x+b y+c z=d\) are parallel lines in the \(x y\) -plane, provided \(a^{2}+b^{2} \neq 0\) and \(c \neq 0\).

Maximizing a sum. Find the maximum value of \(x_{1}+x_{2}+x_{3}+x_{4}\) subject to the condition that \(x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}=16\)

Find the absolute maximum and minimum values of the following functions over the given regions \(R .\) \(f(x, y)=x^{2}+y^{2}-2 y+1 ; R=\left\\{(x, y): x^{2}+y^{2} \leq 4\right\\}\) (This is Exercise \(47, \text { Section } 15.7 .)\)

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