Problem 375 - Calculus Volume 3

Question

Minimize \(f(x, y)=x^{2}+y^{2}, x+2 y-5=0\).

Step-by-Step Solution

Answer

Minimum value is 5 at \((x, y) = (1, 2)\).

Step 1: Understand the Problem

The problem asks us to minimize the function \( f(x, y) = x^2 + y^2 \) subject to the constraint \( x + 2y - 5 = 0 \). This means we're looking for values of \( x \) and \( y \) that satisfy the constraint and give the smallest possible value of \( f(x, y) \).

Step 2: Use the Constraint to Express One Variable

From the constraint \( x + 2y - 5 = 0 \), solve for \( x \) in terms of \( y \). Doing so, we get \( x = 5 - 2y \).

Step 3: Substitute into the Objective Function

Substitute \( x = 5 - 2y \) into the objective function \( f(x, y) = x^2 + y^2 \). This gives us: \[ f(y) = (5 - 2y)^2 + y^2 \]

Step 4: Simplify the New Function

Expand the equation \( f(y) = (5 - 2y)^2 + y^2 \) to simplify: \[ f(y) = (25 - 20y + 4y^2) + y^2 \] \[ f(y) = 5y^2 - 20y + 25 \]

Step 5: Find the Critical Points

Find the derivative of \( f(y) \) with respect to \( y \): \[ f'(y) = 10y - 20 \]Set \( f'(y) = 0 \) to find critical points:\[ 10y - 20 = 0 \]\[ y = 2 \]

Step 6: Solve for Corresponding Value of x

Use \( y = 2 \) in the constraint equation to find \( x \):\[ x = 5 - 2(2) = 1 \]

Step 7: Calculate Minimum Value of the Function

Plug \( x = 1 \) and \( y = 2 \) back into the function:\[ f(1, 2) = 1^2 + 2^2 = 1 + 4 = 5 \]

Step 8: Conclude the Result

The minimum value of the function \( f(x, y) = x^2 + y^2 \) given the constraint \( x + 2y - 5 = 0 \) is 5, and it occurs at \( (x, y) = (1, 2) \).

Key Concepts

Constrained Optimization

Constrained optimization is a fundamental concept in calculus where you seek to optimize a function subject to certain restrictions or constraints.
For example, if we want to minimize the function \(f(x, y) = x^2 + y^2\), we look for the pair \((x, y)\) where \(x + 2y - 5 = 0\) while minimizing \(f(x, y)\). The challenge is to account for both the function we want to minimize and the condition to be satisfied simultaneously.
By using Lagrange multipliers, we can incorporate the constraint into our calculus to find the best solution. Lagrange multipliers are values that allow us to include the constraint directly in derivative equations, making it easier to determine critical points that satisfy both requirements. Remember, constraints are not restrictions; they guide us to the optimal point within a bounded space.

Critical Points

Critical points are key in finding the optimal values of a function, especially in constrained optimization. These are points where the derivative is zero or undefined. In our exercise, finding the critical points of \(f(y)\) involves calculating \(f'(y)\) and setting it to zero.
To find critical points:

• Derive the equation to get \(f'(y)\).
• Set \(f'(y) = 0\) and solve for \(y\).

Once you find \(y\), use the constraints to determine the corresponding \(x\). Critical points can indicate a local maximum, minimum, or a saddle point, hence why further testing might be necessary, especially in more complex functions.

Multivariable Calculus

Multivariable calculus deals with functions depending on two or more variables. In the case of \(f(x, y) = x^2 + y^2\), both \(x\) and \(y\) are involved. Here, we assess how changes in one variable affect the entire function, underlining the importance of derivatives.
Multivariable functions are frequent in optimization problems because real-world applications usually have more than one influencing factor. In problems like these, partial derivatives are useful as they show the effect of changing one variable while keeping others constant. By combining these derivatives with constraints, we navigate the interplay between variables to find an optimum.

Derivative Calculation

Derivative calculation is essential for identifying the nature of functions and finding extrema. For constrained optimization, as seen in our example, you derive the objective function and set its derivative equal to zero.
Steps for derivative calculation include:

• Identify the function you need to differentiate, such as \(f(y)\).
• Use differentiation techniques to compute \(f'(y)\).
• Simplify the derivative, then solve the equation obtained by setting it zero to find potential extremes.

Once you determine where the derivative is zero, you've found critical points—potential players in your optimizations. Derivatives also help analyze the curve's behavior around these points, hinting whether these points correspond to minima or maxima.