Problem 372 - Calculus Volume 3

Question

$$ \begin{aligned} f(x, y) &=x^{2}-y^{2} ; x>0, y>0 ; \\ \text { Maximize } g(x, y) &=y-x^{2}=0 \end{aligned} $$

Step-by-Step Solution

Answer

The maximum value of \( f(x, y) \) is \( \frac{1}{4} \) at \( x = \frac{1}{\sqrt{2}}, y = \frac{1}{2} \).

Step 1: Understanding the Problem

We need to maximize the function \( f(x, y) = x^2 - y^2 \) subject to the constraint \( g(x, y) = y - x^2 = 0 \). This means \( y = x^2 \).

Step 2: Substitute Constraint into Function

Substitute the constraint \( y = x^2 \) into the function: \( f(x, y) = f(x, x^2) = x^2 - (x^2)^2 = x^2 - x^4 \).

Step 3: Differentiate the Substituted Function

Calculate the derivative of \( f(x, x^2) = x^2 - x^4 \) with respect to \( x \). We have:\[ \frac{d}{dx}(x^2 - x^4) = 2x - 4x^3. \]

Step 4: Set the Derivative to Zero

To find critical points, set the derivative equal to zero:\[ 2x - 4x^3 = 0 \]Factor:\[ 2x(1 - 2x^2) = 0. \]

Step 5: Solve for Critical Points

Solve \( 2x(1 - 2x^2) = 0 \). This gives us \( x = 0 \) or \( 1 - 2x^2 = 0 \). The latter gives \( x^2 = \frac{1}{2} \), so \( x = \frac{1}{\sqrt{2}} \) (since \( x > 0 \)).

Step 6: Calculate Corresponding y-values

Using \( y = x^2 \), calculate the corresponding \( y \)-values. For \( x = \frac{1}{\sqrt{2}} \), we get \( y = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2} \).

Step 7: Evaluate Function at Critical Point

Evaluate \( f(x, y) = x^2 - y^2 \) at \( x = \frac{1}{\sqrt{2}}, y = \frac{1}{2} \):\[ f\left(\frac{1}{\sqrt{2}}, \frac{1}{2}\right) = \left(\frac{1}{\sqrt{2}}\right)^2 - \left(\frac{1}{2}\right)^2 = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}. \]

Step 8: Confirm Maximum

Since the function \( x^2 - x^4 \) can have only one critical point for \( x > 0 \), this point is a maximum given the constraint.

Key Concepts

Critical Points

Critical points play a fundamental role in optimization problems. They are values where the derivative of a function is zero or undefined. These points are key because they indicate potential maxima, minima, or saddle points of the function. In the context of the problem, after substituting the constraint and forming the new function, we calculate the derivative and find where it equals zero. This step helps identify possible locations of optimum values within the constraint boundaries.
For this problem, after differentiating the function, the equation obtained is \( 2x - 4x^3 = 0 \). To solve it, set each factor to zero: \( 2x = 0 \) (which is discarded since \( x > 0 \)), and \( 1 - 2x^2 = 0 \). Solving this provides \( x = \frac{1}{\sqrt{2}} \). This critical point is then used to find corresponding \( y \)-values, giving a complete coordinate for potential maxima or minima.

Differentiation

Differentiation is a core concept in calculus used for finding the rate at which a function is changing. In optimization problems, it helps us determine critical points by calculating where the function's derivative is zero.
For example, for the substituted function \( f(x, x^2) = x^2 - x^4 \), differentiation with respect to \( x \) leads to the derivative \( 2x - 4x^3 \). By setting this to zero, you're identifying the points where the slope of the function’s tangent is zero. These points are essential for locating potential peaks or valleys in the function and are thus crucial for constrained optimization problems.

Constraint Substitution

Constraint substitution is a technique used in constrained optimization problems where the constraint equation is substituted directly into the objective function. This simplification reduces the number of variables, making it easier to perform calculations.
In this exercise, we start with the constraint \( g(x, y) = y - x^2 = 0 \), or \( y = x^2 \). Substituting \( y = x^2 \) into the objective function \( f(x, y) = x^2 - y^2 \) provides a new function depending solely on \( x \): \( f(x, x^2) = x^2 - x^4 \). After substitution, the problem simplifies significantly, focusing only on one variable, making the subsequent differentiation and search for critical points much more manageable.