Problem 4
Question
Find the extremum of \(f(x, y)\) subject to the given constraint, and state whether it is a maximum or a minimum. $$ f(x, y)=x^{2}+y^{2} ; x+4 y=17 $$
Step-by-Step Solution
Verified Answer
The minimum occurs at \((1, 4)\).
1Step 1: Identify the Objective and Constraint
We need to find the extremum (maximum or minimum) of the function \( f(x, y) = x^2 + y^2 \) subject to the constraint \( x + 4y = 17 \).
2Step 2: Apply Lagrange Multipliers
Form the Lagrangian, which is \( \mathcal{L}(x, y, \lambda) = x^2 + y^2 + \lambda(x + 4y - 17) \). Our goal is to solve \( abla \mathcal{L} = 0 \).
3Step 3: Compute Partial Derivatives
The partial derivatives are:\[ \frac{\partial \mathcal{L}}{\partial x} = 2x + \lambda = 0 \]\[ \frac{\partial \mathcal{L}}{\partial y} = 2y + 4\lambda = 0 \]\[ \frac{\partial \mathcal{L}}{\partial \lambda} = x + 4y - 17 = 0 \]
4Step 4: Solve the System of Equations
From \( 2x + \lambda = 0 \), we get \( \lambda = -2x \). From \( 2y + 4\lambda = 0 \), solve for \( \lambda \):\[ 4\lambda = -2y \Rightarrow \lambda = -\frac{y}{2} \]Setting the two expressions for \( \lambda \) equal, we have:\[ -2x = -\frac{y}{2} \Rightarrow y = 4x \].
5Step 5: Substitute into the Constraint
Substitute \( y = 4x \) into \( x + 4y = 17 \):\[ x + 4(4x) = 17 \Rightarrow x + 16x = 17 \Rightarrow 17x = 17 \Rightarrow x = 1 \]Substituting back for \( y \), we get \( y = 4(1) = 4 \).
6Step 6: Determine the Nature of the Extremum
Since \( f(x, y) = x^2 + y^2 \) is a sum of squares, it always increases away from the origin. Thus, the minimum value happens at \( x = 1 \) and \( y = 4 \), the only critical point we found. Thus it is a minimum.
Key Concepts
ExtremumConstraint OptimizationPartial Derivatives
Extremum
An extremum refers to either the maximum or minimum value of a function. In the context of mathematics, identifying extrema helps us understand optimization problems.
For the given function, we aim to find out whether the function value is increasing or decreasing, thereby helping us recognize if there's a peak or a trough. Here we deal explicitly with potential minima or maxima.
To conclude whether a point grants either of these values, additional tests or the nature of the function (as seen in the exercise) are considered.
For the given function, we aim to find out whether the function value is increasing or decreasing, thereby helping us recognize if there's a peak or a trough. Here we deal explicitly with potential minima or maxima.
To conclude whether a point grants either of these values, additional tests or the nature of the function (as seen in the exercise) are considered.
- If the function tends to increase as we move away from a point, it's a minimum.
- If it decreases, it's a maximum.
Constraint Optimization
Constraint Optimization is a technique used when optimizing a function subject to specific limitations or conditions.
In this exercise, our task was to find the extremum of function \(f(x, y) = x^2 + y^2\) with a constraint \(x + 4y = 17\). Constraints bind the solution space to a feasible region.
Approaches for constraint optimization are varied but one popular method is using Lagrange multipliers. This involves augmenting the original function with components for constraints using additional variables.
When constraints are linear, they can be visualized as straight lines cutting through the function's graph. They limit the potential solution points, concentrating analysis on intersection points of these constraints with the function's derivatives.
In this exercise, our task was to find the extremum of function \(f(x, y) = x^2 + y^2\) with a constraint \(x + 4y = 17\). Constraints bind the solution space to a feasible region.
Approaches for constraint optimization are varied but one popular method is using Lagrange multipliers. This involves augmenting the original function with components for constraints using additional variables.
When constraints are linear, they can be visualized as straight lines cutting through the function's graph. They limit the potential solution points, concentrating analysis on intersection points of these constraints with the function's derivatives.
Partial Derivatives
Partial derivatives are crucial in understanding the rate at which a function changes with respect to its variables individually, keeping others constant.
For multivariable functions like \(f(x, y)\), derivatives with respect to \(x\) (\(\frac{\partial f}{\partial x}\)) and \(y\) (\(\frac{\partial f}{\partial y}\)) depict how the function surface shifts over these axes.
For multivariable functions like \(f(x, y)\), derivatives with respect to \(x\) (\(\frac{\partial f}{\partial x}\)) and \(y\) (\(\frac{\partial f}{\partial y}\)) depict how the function surface shifts over these axes.
- A zero partial derivative suggests a local extremum point since perturbations in the variable lead to no immediate slope change.
- In the exercise, the concept underpins the establishment of our system of equations to find critical points.
Other exercises in this chapter
Problem 4
Evaluate. $$ \int_{1}^{4} \int_{-2}^{1} x^{3} y d y d x $$
View solution Problem 4
Find \(\frac{\partial z}{\partial x}, \frac{\partial z}{\partial y},\left.\frac{\partial z}{\partial x}\right|_{(-2,-3)},\) and \(\left.\frac{\partial z}{\parti
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Find the regression line for each data set. $$ \begin{array}{|c|c|c|c|} \hline x & 1 & 2 & 4 \\ \hline y & 3 & 5 & 8 \\ \hline \end{array} $$
View solution Problem 4
Find the relative maximum and minimum values. $$ f(x, y)=4 x y-x^{3}-2 y^{2} $$
View solution