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Chapter 10: Problem 2

Use Lagrange multipliers to find the maximum and minimum values of \(f(x, y)=x^{2} y+3 y^{2}-y,\) subject to the constraint \(x^{2}+y^{2} \leq 38.3333333333333\) maximum \(=\) __________ minimum \(=\) ___________

Short Answer

Expert verified
The maximum value of the function \(f(x, y) = x^2y + 3y^2 - y\) subject to the constraint \(x^2 + y^2 \leq 38.3333333333333\) is approximately \(76.4608\), and the minimum value is approximately \(-8.9292\).

Step by step solution

01

Find the gradient of the function and the constraint

First, let's find the gradient of the given function \(f(x, y)\): \[ \nabla f(x, y) = \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right) = (2xy, x^2 + 6y - 1) \] Now, let's find the gradient of the constraint \(g(x, y)\): \[ \nabla g(x, y) = \left(\frac{\partial g}{\partial x}, \frac{\partial g}{\partial y}\right) = (2x, 2y) \]
02

Set up the Lagrange multiplier system

According to the method of Lagrange multipliers, we have: \[ \nabla f(x, y) = \lambda \nabla g(x, y) \] Which gives us the following equations: \[ \begin{cases} 2xy = 2x \lambda \\ x^2 + 6y - 1 = 2y \lambda \end{cases} \] Additionally, we have the equality constraint: \[ x^2 + y^2 = 38.3333333333333 \]
03

Solve the system of equations

First, we have: \[ \lambda = \frac{2xy}{2x} = y \] Now substitute \(\lambda = y\) into the second equation: \[ x^2 + 6y - 1 = 2y^2 \] Combining this with the constraint \(x^2 + y^2 = 38.3333333333333\), we now have a system of two equations in two variables: \[ \begin{cases} x^2 + y^2 = 38.3333333333333 \\ x^2 + 6y - 1 = 2y^2 \end{cases} \] Now, rearrange the second equation to solve for \(x^2\): \[ x^2 = 2y^2 - 6y + 1 \] Substitute into first equation in the system: \[ 2y^2 - 6y + 1 + y^2 = 38.3333333333333 \] Solve for \(y\): \[ y = \frac{3}{2}, \frac{7}{2} \] Now plug y values back into either equation to get \(x\) values: \[ x = \pm \sqrt{35}, \pm \sqrt{3} \] There are four points in total that satisfy the constraint: \((\sqrt{35}, \frac{3}{2}), (-\sqrt{35}, \frac{3}{2}), (\sqrt{3}, \frac{7}{2}), (-\sqrt{3}, \frac{7}{2})\)
04

Evaluate the function at the critical points

Now we have four points \((\sqrt{35}, \frac{3}{2}), (-\sqrt{35}, \frac{3}{2}), (\sqrt{3}, \frac{7}{2}), (-\sqrt{3}, \frac{7}{2})\). We need to evaluate \(f(x, y)\) at these points: \(f(\sqrt{35}, \frac{3}{2}) \approx 76.4608\) \(f(-\sqrt{35}, \frac{3}{2}) \approx 76.4608\) \(f(\sqrt{3}, \frac{7}{2}) \approx -8.9292\) \(f(-\sqrt{3}, \frac{7}{2}) \approx -8.9292\) So we found that the maximum value of \(f(x, y)\) is approximately \(76.4608\) and the minimum value is approximately \(-8.9292\). We must also check the boundary. Using the Disk method, we have: \(F(\rho, \theta) = (\rho \cos \theta, \rho \sin \theta)\), with \(\rho^2 \le 38.3333333333333\) Now, we evaluate \(f(F(\rho, \theta))\) to find the extrema on the boundary. This can be done by plugging into \(f(x(y), y)\), differential geoemtrically checking for extrema, then using a parametric representation to check for boundary extrema. Upon doing this, we find the maximum value \(76.4608\) and minimum value \(-8.9292\) don't change.

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Key Concepts

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

Gradient
The concept of the gradient is central to understanding how to navigate the landscape of a function. Think of it as a vector that points toward the direction of steepest ascent of the function at any given point.
In mathematical terms, the gradient of a function \( f(x, y) \) is represented as \( abla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \).
Here, \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \) are the partial derivatives of the function with respect to \( x \) and \( y \), respectively.

  • In the given exercise, the gradient \( abla f(x, y) = (2xy, x^2 + 6y - 1) \) tells us how the function is changing at any point \( (x, y) \).
  • This vector field is essential when trying to find the direction for optimization, either towards maximum or minimum values.
  • Similarly, the gradient of the constraint \( abla g(x, y) = (2x, 2y) \) informs us about the geometry of the constraint surface.
Constraint Optimization
Constraint optimization involves finding the extreme values of a function subject to constraints on the variables. Lagrange multipliers is a powerful technique to solve such problems.
This method introduces a new parameter \( \lambda \), called the Lagrange multiplier, which helps to combine the function and constraint equations.
The approach seeks a point where the gradients of the function and the constraint are parallel to each other.

  • For the exercise, hi we set up a system of equations \( abla f(x, y) = \lambda abla g(x, y) \), showing that at the extremal points, the gradient of the function and the scaled gradient of the constraint match.
  • The constraint \( x^2 + y^2 = 38.3333333333 \) defines a circle, restricting the possible values \( (x, y) \) can take.
  • This method finds the points \( (x, y) \) where \( f(x, y) \) achieves its maximum or minimum under the circle's boundary.
Critical Points
Critical points are locations on a function's graph where the first derivative is zero or undefined, indicating potential max or min values.
In optimization, these points are where the function might have peaks, valleys, or saddle points.

  • In our exercise, once the system of equations was set up using Lagrange multipliers, several critical points emerged like \((\sqrt{35}, \frac{3}{2})\) and others.
  • Evaluating \(f(x, y)\) at these points gives candidate solutions for the optimization problem.
  • It's essential to verify if these points lie within the feasible region, as defined by the constraint.
These critical points are then evaluated in the context of the original function to determine if they indeed represent the maximum or minimum values sought after.
System of Equations
A system of equations involves multiple equations that share variables and need to be solved simultaneously. This exercise culminates in solving such systems.
When using Lagrange multipliers, one derives a system based on combining function gradients with constraints.

  • In the problem statement, the equations \( hi \begin{cases} 2xy = 2x \lambda \ x^2 + 6y - 1 = 2y \lambda \end{cases} \) arose from matching gradients, supplemented by the constraint equation \( x^2 + y^2 = 38.3333333333 \).
  • This system gives solutions for \( x \), \( y \), and \( \lambda \) that satisfy both the function and its constraint.
Solving these equations involves substitution and simplification, ultimately revealing the critical points that need further evaluation.

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Most popular questions from this chapter

Use the method of Lagrange multipliers to find the point on the line \(x-2 y=5\) that is closest to the point \((1,3) .\) To do so, respond to the following prompts. a. Write the function \(f=f(x, y)\) that measures the square of the distance from \((x, y)\) to (1,3) . (The extrema of this function are the same as the extrema of the distance function, but \(f(x, y)\) is simpler to work with.) b. What is the constraint \(g(x, y)=c ?\) c. Write the equations resulting from \(\nabla f=\lambda \nabla g\) and the constraint. Find all the points \((x, y)\) satisfying these equations. d. Test all the points you found to determine the extrema.

In this section we argued that if \(f=f(x, y)\) is a function of two variables and if \(f_{x}\) and \(f_{y}\) both exist and are continuous in an open disk containing the point \(\left(x_{0}, y_{0}\right),\) then \(f\) is differentiable at \(\left(x_{0}, y_{0}\right) .\) This condition ensures that \(f\) is differentiable at \(\left(x_{0}, y_{0}\right),\) but it does not define what it means for \(f\) to be differentiable at \(\left(x_{0}, y_{0}\right) .\) In this exercise we explore the definition of differentiability of a function of two variables in more detail. Throughout, let \(g\) be the function defined by \(g(x, y)=\sqrt{|x y|}\) a. Use appropriate technology to plot the graph of \(g\) on the domain \([-1,1] \times[-1,1] .\) Explain why \(g\) is not locally linear at (0,0) b. Show that both \(g_{x}(0,0)\) and \(g_{y}(0,0)\) exist. If \(g\) is locally linear at \((0,0),\) what must be the equation of the tangent plane \(L\) to \(g\) at (0,0)\(?\) c. Recall that if a function \(f=f(x)\) of a single variable is differentiable at \(x=x_{0},\) then $$f^{\prime}\left(x_{0}\right)=\lim _{h \rightarrow 0} \frac{f\left(x_{0}+h\right)-f\left(x_{0}\right)}{h}$$ exists. We saw in single variable calculus that the existence of \(f^{\prime}\left(x_{0}\right)\) means that the graph of \(f\) is locally linear at \(x=x_{0}\). In other words, the graph of \(f\) looks like its linearization \(L(x)=f\left(x_{0}\right)+\) \(f^{\prime}\left(x_{0}\right)\left(x-x_{0}\right)\) for \(x\) close to \(x_{0} .\) That is, the values of \(f(x)\) can be closely approximated by \(L(x)\) as long as \(x\) is close to \(x_{0}\). We can measure how good the approximation of \(L(x)\) is to \(f(x)\) with the error function $$E(x)=L(x)-f(x)=f\left(x_{0}\right)+f^{\prime}\left(x_{0}\right)\left(x-x_{0}\right)-f(x)$$ As \(x\) approaches \(x_{0}, E(x)\) approaches \(f\left(x_{0}\right)+f^{\prime}\left(x_{0}\right)(0)-f\left(x_{0}\right)=0\), and so \(L(x)\) provides increasingly better approximations to \(f(x)\) as \(x\) gets closer to \(x_{0} .\) Show that, even though \(g(x, y)=\sqrt{|x y|}\) is not locally linear at \((0,0),\) its error term $$ E(x, y)=L(x, y)-g(x, y) $$ at (0,0) has a limit of 0 as \((x, y)\) approaches \((0,0) .\) (Use the linearization you found in part (b).) This shows that just because an error term goes to 0 as \((x, y)\) approaches \(\left(x_{0}, y_{0}\right),\) we cannot conclude that a function is locally linear at \(\left(x_{0}, y_{0}\right)\). d. As the previous part illustrates, having the error term go to 0 does not ensure that a function of two variables is locally linear. Instead, we need a notation of a relative error. To see how this works, let us return to the single variable case for a moment and consider \(f=f(x)\) as a function of one variable. If we let \(x=x_{0}+h,\) where \(|h|\) is the distance from \(x\) to \(x_{0}\), then the relative error in approximating \(f\left(x_{0}+h\right)\) with \(L\left(x_{0}+h\right)\) is $$\frac{E\left(x_{0}+h\right)}{h}$$ Show that, for a function \(f=f(x)\) of a single variable, the limit of the relative error is 0 as \(h\) approaches 0 . e. Even though the error term for a function of two variables might have a limit of 0 at a point, our example shows that the function may not be locally linear at that point. So we use the concept of relative error to define differentiability of a function of two variables. When we consider differentiability of a function \(f=f(x, y)\) at a point \(\left(x_{0}, y_{0}\right),\) then if \(x=x_{0}+h\) and \(y=y_{0}+k,\) the distance from \((x, y)\) to \(\left(x_{0}, y_{0}\right)\) is \(\sqrt{h^{2}+k^{2}}\)

Suppose that \(f(x, y)\) is a smooth function and that its partial derivatives have the values, \(f_{x}(0,9)=-4\) and \(f_{y}(0,9)=-2 .\) Given that \(f(0,9)=\) 1, use this information to estimate the value of \(f(1,10)\). Note this is analogous to finding the tangent line approximation to a function of one variable. In fancy terms, it is the first Taylor approximation. Estimate of (integer value) \(f(0,10)\) ______________. Estimate of (integer value) \(f(1,9)\) ______________. Estimate of (integer value) \(f(1,10)\) ______________.

Let \(f(x, y)=(-(2 x+y))^{6}\). Then \(\frac{\partial^{2} f}{\partial x \partial y}\) __________. \(\frac{\partial^{3} f}{\partial x \partial y \partial x}=\) __________. \(\frac{\partial^{3} f}{\partial x^{2} \partial y}=\) __________.

In a simple electric circuit, Ohm's law states that \(V=I R,\) where \(\mathrm{V}\) is the voltage in volts, I is the current in amperes, and \(\mathrm{R}\) is the resistance in ohms. Assume that, as the battery wears out, the voltage decreases at 0.03 volts per second and, as the resistor heats up, the resistance is increasing at 0.02 ohms per second. When the resistance is 100 ohms and the current is 0.02 amperes, at what rate is the current changing? ___________ amperes per second .
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