Problem 10 - Calculus and its applications

Question

Find the relative maximum and minimum values. $$ f(x, y)=4 y+6 x-x^{2}-y^{2} $$

Step-by-Step Solution

Answer

The function has a relative maximum at \((3, 2)\), but no relative minimum.

Step 1: Find the Critical Points

To find the critical points of the function, calculate the partial derivatives of the function \(f(x, y) = 4y + 6x - x^2 - y^2\):\[ \frac{\partial f}{\partial x} = 6 - 2x \] and \[ \frac{\partial f}{\partial y} = 4 - 2y. \] Set these partial derivatives equal to zero to find the critical points. Solve \(6 - 2x = 0\) to get \(x = 3\) and solve \(4 - 2y = 0\) to get \(y = 2\). Thus, the critical point is \((3, 2)\).

Step 2: Use the Second Derivative Test

Now, apply the second derivative test to classify the critical point. Calculate the second partial derivatives: \[ \frac{\partial^2 f}{\partial x^2} = -2, \] \[ \frac{\partial^2 f}{\partial y^2} = -2, \] and the mixed second partial derivative \( \frac{\partial^2 f}{\partial x \partial y} = 0 \). The determinant of the Hessian matrix is: \[ D = \left( \frac{\partial^2 f}{\partial x^2} \right) \left( \frac{\partial^2 f}{\partial y^2} \right) - \left( \frac{\partial^2 f}{\partial x \partial y} \right)^2 = (-2)(-2) - 0^2 = 4. \] Since \(D > 0\) and \( \frac{\partial^2 f}{\partial x^2} < 0\), the point \((3, 2)\) is a relative maximum.

Step 3: Conclusion on Relative Minima

Since the Hessian determinant is positive and the second derivative of \(\frac{\partial^2 f}{\partial x^2}\) is negative, all inferences point to the fact that there exists no relative minimum in the region we have solved. The critical point found is a relative maximum. Consequently, no relative minimum can be derived from the second derivative's conditions at this point.

Key Concepts

Critical Points

In multivariable calculus, finding critical points is essential for determining where a function has potential extrema (maxima or minima). A critical point occurs when the partial derivatives of a function are zero, indicating a horizontal tangent plane. In simpler terms, the slope in all directions is zero.
To locate the critical points of a function like \( f(x, y) = 4y + 6x - x^2 - y^2 \), we start by computing its partial derivatives:

• The partial derivative with respect to \( x \) is \( \frac{\partial f}{\partial x} = 6 - 2x \).
• The partial derivative with respect to \( y \) is \( \frac{\partial f}{\partial y} = 4 - 2y \).

By setting these partial derivatives to zero, \( 6 - 2x = 0 \) and \( 4 - 2y = 0 \), we solve for \( x \) and \( y \). This gives us the critical point \((3, 2)\). Here, both slopes (or changes) in the \(x\) and \(y\) directions are zero, making it a candidate for a relative extremum.

Second Derivative Test

Determining the nature of critical points further involves the second derivative test. This test reveals whether the critical point is a minimum, maximum, or a saddle point. The test uses the second partial derivatives to form a Hessian matrix and its determinant, which provides crucial insight.
For the function given, we compute:

• \( \frac{\partial^2 f}{\partial x^2} = -2 \)
• \( \frac{\partial^2 f}{\partial y^2} = -2 \)
• The mixed second partial derivative \( \frac{\partial^2 f}{\partial x \partial y} = 0 \)

The determinant of the Hessian matrix \( D \) is calculated as:\[ D = \left( \frac{\partial^2 f}{\partial x^2} \right) \left( \frac{\partial^2 f}{\partial y^2} \right) - \left( \frac{\partial^2 f}{\partial x \partial y} \right)^2 = (-2)(-2) - 0^2 = 4 \]
Since \( D > 0 \) and \( \frac{\partial^2 f}{\partial x^2} < 0 \), the critical point \((3, 2)\) is identified as a relative maximum. The negative value of \( \frac{\partial^2 f}{\partial x^2} \) confirms that the function curves downward at this point.

Partial Derivatives

Partial derivatives are a fundamental tool in multivariable calculus used to analyze functions of several variables. A partial derivative measures how a function changes as only one variable changes, while all others remain constant.
In our exercise, the function \( f(x, y) = 4y + 6x - x^2 - y^2 \) has two variables, \( x \) and \( y \). To find the critical points, we first take the partial derivatives with respect to \( x \) and \( y \). This is achieved by differentiating the function while treating the other variable as a constant:

• For \( x \), this results in \( \frac{\partial f}{\partial x} = 6 - 2x \).
• For \( y \), this results in \( \frac{\partial f}{\partial y} = 4 - 2y \).

Setting these derivatives to zero helps in discovering critical points where the function does not change instantaneously in any direction. Understanding partial derivatives is crucial for analyzing how a function behaves and for pinpointing critical points in the context of optimizing and graphing functions in higher dimensions.