Problem 20
Question
Consider the problem of maximizing the function \(f(x, y)=2 x+3 y\) subject to the constraint \(\sqrt{x}+\sqrt{y}=5\) (a) Try using Lagrange multipliers to solve the problem. (b) Does \(f(25,0)\) give a larger value than the one in part (a)? (c) Solve the problem by graphing the constraint equation and several level curves of \(f\) . (d) Explain why the method of Lagrange multipliers fails to solve the problem. (e) What is the significance of \(f(9,4) ?\)
Step-by-Step Solution
Verified Answer
The largest value from the constraint is 50, at (25,0), despite (9,4) being the constrained maximum.
1Step 1: Understanding the Lagrange Multiplier Method
The method of Lagrange multipliers is used to find the maxima and minima of a function subject to a constraint. We introduce a Lagrange multiplier \( \lambda \) and set up the equations: \( abla f = \lambda abla g \) where \( g(x, y) = \sqrt{x} + \sqrt{y} - 5 = 0 \).
2Step 2: Applying the Lagrange Multiplier Method
Compute the gradients: \( abla f = (2, 3) \) and \( abla g = \left( \frac{1}{2\sqrt{x}}, \frac{1}{2\sqrt{y}} \right) \). Set \( 2 = \lambda \frac{1}{2\sqrt{x}} \) and \( 3 = \lambda \frac{1}{2\sqrt{y}} \). This gives two equations: \( 4\sqrt{x} = \lambda \) and \( 6\sqrt{y} = \lambda \). Equating these, we have \( 4\sqrt{x} = 6\sqrt{y} \).
3Step 3: Solving for Critical Points
From \( 4\sqrt{x} = 6\sqrt{y} \), we have \( \frac{\sqrt{x}}{\sqrt{y}} = \frac{3}{2} \). Squaring both sides gives \( \frac{x}{y} = \frac{9}{4} \). Substitute \( y = \frac{4}{9}x \) into the constraint \( \sqrt{x} + \sqrt{y} = 5 \) to get \( \sqrt{x} + \frac{2}{3}\sqrt{x} = 5 \). This simplifies to \( \frac{5}{3}\sqrt{x} = 5 \), thus \( \sqrt{x} = 3 \), so \( x = 9 \). Substituting back, \( y = 4 \).
4Step 4: Evaluating the Constraint Solution
Substitute \( x = 9 \) and \( y = 4 \) into the function \( f(x, y) = 2x + 3y \): \( f(9, 4) = 2(9) + 3(4) = 18 + 12 = 30 \).
5Step 5: Comparing with Given Points
Evaluate \( f(25, 0) = 2(25) + 3(0) = 50 \). Since 50 is greater than 30, \( f(25, 0) \) gives a larger value.
6Step 6: Graphical Interpretation
The constraint \( \sqrt{x} + \sqrt{y} = 5 \) represents a curve. By graphing the level curves of \( f(x, y) = 2x + 3y \) and identifying which level curve is tangent to the constraint, we find \( (9,4) \) is a solution where the constraint touches the level curve.
7Step 7: Explanation of Lagrange Multiplier Failure
The Lagrange multiplier method fails when the constraint curve is not differentiable at the optimal point or the gradients do not define a tangent due to singularities such as \( x \) or \( y \) being zero.
8Step 8: Significance of \(f(9,4)\)
\( f(9,4) = 30 \) represents the constrained maximum obtained via solving Lagrange multipliers, but it is not globally optimal owing to specific values that provide higher outputs.
Key Concepts
MaximizationConstraint OptimizationLevel CurvesGradient Method
Maximization
Maximization is the process of finding the highest or most extreme value of a function. In this problem, we aim to maximize the function \( f(x, y) = 2x + 3y \). The goal is to find the point \((x, y)\) where this function has its largest possible value, considering the given constraint. This point can be one where certain conditions are met, often at the boundary of the allowed region defined by the constraint.
In the problem given, we see this with the calculation of \( f(9, 4) \), which is found to be 30. While this exercise involved finding a maximum under a specific constraint, it's important to note that maxima can be absolute (global) or relative (local). In this case, \( f(25, 0) = 50 \) shows a higher value, indicating that the unconstrained maximum might be higher than the constrained maximum.
In the problem given, we see this with the calculation of \( f(9, 4) \), which is found to be 30. While this exercise involved finding a maximum under a specific constraint, it's important to note that maxima can be absolute (global) or relative (local). In this case, \( f(25, 0) = 50 \) shows a higher value, indicating that the unconstrained maximum might be higher than the constrained maximum.
Constraint Optimization
Constraint Optimization is about maximizing or minimizing a function subject to constraints that restrict the domain. This defined constraint often represents a real-world limitation, such as the total resources available or a fixed supply.
In this particular problem, the constraint equation \( \sqrt{x} + \sqrt{y} = 5 \) describes a limitation where the sum of the square roots of \( x \) and \( y \) must always equal 5. The objective here is to find where within this constraint the function \( f(x, y) \) can achieve its largest value. Understanding the interplay between the function and the constraint is a key component of constraint optimization. In practice, this might involve trade-offs where certain variables can be increased only at the expense of decreasing others, all while honoring the constraint guideline.
In this particular problem, the constraint equation \( \sqrt{x} + \sqrt{y} = 5 \) describes a limitation where the sum of the square roots of \( x \) and \( y \) must always equal 5. The objective here is to find where within this constraint the function \( f(x, y) \) can achieve its largest value. Understanding the interplay between the function and the constraint is a key component of constraint optimization. In practice, this might involve trade-offs where certain variables can be increased only at the expense of decreasing others, all while honoring the constraint guideline.
Level Curves
Level Curves are the paths formed by plotting points \((x, y)\) that give the same function value \( c = f(x, y) \). These curves help in visualizing the properties of the function over a plane. They are extremely helpful in identifying optimum points when solving optimization problems graphically.
For this function \( f(x, y) = 2x + 3y \), each level curve is a straight line, because \( 2x + 3y = c \). Different values of \( c \) will give us different lines. By analyzing these curves, we can see which one tangentially touches the constraint curve \( \sqrt{x} + \sqrt{y} = 5 \). The point of tangency represents the optimal solution under the given constraint. This tangency visualizes the balance or trade-off imposed by the constraint, aligning it with the direction of the maximum increase of the function.
For this function \( f(x, y) = 2x + 3y \), each level curve is a straight line, because \( 2x + 3y = c \). Different values of \( c \) will give us different lines. By analyzing these curves, we can see which one tangentially touches the constraint curve \( \sqrt{x} + \sqrt{y} = 5 \). The point of tangency represents the optimal solution under the given constraint. This tangency visualizes the balance or trade-off imposed by the constraint, aligning it with the direction of the maximum increase of the function.
Gradient Method
The Gradient Method connects to the idea of using derivatives to determine the direction and rate of steepest ascent for a function. In optimization, gradients show how a function changes as its input changes.
In the Lagrange multiplier method, gradients are crucial. Here, we set the gradient of the function \( abla f = (2, 3) \) equal to a scaled version (via \( \lambda \), the Lagrange multiplier) of the gradient of the constraint \( abla g = \left( \frac{1}{2\sqrt{x}}, \frac{1}{2\sqrt{y}} \right) \). The equations \( 2 = \lambda \frac{1}{2\sqrt{x}} \) and \( 3 = \lambda \frac{1}{2\sqrt{y}} \) arise from aligning these gradients. This technique, although very powerful, fails when the gradients cannot describe the tangency of the level and constraint curves, which is one of the shortcomings highlighted in this textbook problem. In situations where variables approach zero, singularities can disrupt the straightforward application of this method.
In the Lagrange multiplier method, gradients are crucial. Here, we set the gradient of the function \( abla f = (2, 3) \) equal to a scaled version (via \( \lambda \), the Lagrange multiplier) of the gradient of the constraint \( abla g = \left( \frac{1}{2\sqrt{x}}, \frac{1}{2\sqrt{y}} \right) \). The equations \( 2 = \lambda \frac{1}{2\sqrt{x}} \) and \( 3 = \lambda \frac{1}{2\sqrt{y}} \) arise from aligning these gradients. This technique, although very powerful, fails when the gradients cannot describe the tangency of the level and constraint curves, which is one of the shortcomings highlighted in this textbook problem. In situations where variables approach zero, singularities can disrupt the straightforward application of this method.
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