Problem 17 - Applied Calculus

Question

For \(f(x, y)=A-\left(x^{2}+B x+y^{2}+C y\right),\) what values of \(A, B,\) and \(C\) give \(f\) a local maximum value of 15 at the point (-2,1)\(?\)

Step-by-Step Solution

Answer

A = 10, B = 4, C = -2.

Step 1: Substitute Values into the Function

We need to substitute the given point \((-2,1)\) and the local maximum value \(15\) into the function \(f(x, y) = A - (x^{2} + Bx + y^{2} + Cy)\). If the local maximum is at this point, then \(f(-2, 1) = 15\). Therefore, substitute \(x = -2\) and \(y = 1\):\[\ 15 = A - ((-2)^2 + B(-2) + (1)^2 + C(1)). \] This simplifies to:\[\ A - (4 - 2B + 1 + C) = 15. \] Further simplification gives us the equation: \[ A - (5 - 2B + C) = 15. \]

Step 2: Simplify and Solve for A

From the equation obtained from Step 1, simplify and solve for \(A\):\[ A - 5 + 2B - C = 15. \] Rearrange to get: \[ A = 20 - 2B + C. \] This is the expression for \(A\) in terms of \(B\) and \(C\).

Step 3: Calculate Partial Derivatives for Critical Points

To find critical points, compute the partial derivatives with respect to \(x\) and \(y\) and set them equal to zero.\[\ \frac{\partial f}{\partial x} = -2x - B = 0,\] and \[\frac{\partial f}{\partial y} = -2y - C = 0.\] For the point \((-2,1)\), substitute into these equations:\[\ -2(-2) - B = 0 \quad \Rightarrow \quad 4 = B,\] and \[ -2(1) - C = 0 \quad \Rightarrow \quad C = -2. \] This gives us \(B = 4\) and \(C = -2\).

Step 4: Substitute B and C Values Back

Now substitute \(B = 4\) and \(C = -2\) into the equation found in Step 2 for \(A\):\[ A = 20 - 2(4) + (-2). \] Calculate this to get \[ A = 20 - 8 - 2 = 10. \] Thus, \(A = 10\).

Key Concepts

Critical Points

Critical points in multivariable calculus are the points where the gradient of the function is zero or undefined. This means the slope of the tangent plane to the surface at those points is flat, making it potential candidates for local maximum, minimum, or saddle points. To find the critical points of a function, we typically find the partial derivatives with respect to each variable and set them equal to zero.

• The gradient \( abla f(x, y) \) consists of partial derivatives \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \).
• Setting \( abla f(x, y) = 0 \) helps locate critical points like in our exercise with \( f(x, y) = A - (x^2 + Bx + y^2 + Cy) \).
• For the function \( f(x, y) = 15 \) at the point \( (-2, 1) \), critical points are found by solving: \( -2x - B = 0 \) and \( -2y - C = 0 \).

Understanding critical points is essential because they help calculate where a function's behavior changes or reaches specific values, such as a local maximum or minimum.

Local Maximum

A local maximum occurs in a function when a point is higher than all other nearby points. Unlike a global maximum, a local maximum may not be the highest point on the entire graph but is the highest compared to its immediate surroundings.
The local maximum condition can be verified using the second partial derivatives test. For a function of two variables, this involves computing the determinant of the Hessian matrix, which is derived from the second partial derivatives.

• A local maximum is assumed at the critical point if the determinant of the Hessian is positive and the second partial derivatives are negative.
• In our exercise, the given function achieves a local maximum of 15 at the point \( (-2, 1) \).

Local maxima are crucial in various applications as they help in optimization problems and understanding the behavior of real-world phenomena.

Partial Derivatives

Partial derivatives are foundational to understanding multivariable functions. They measure how the function changes as each variable changes while keeping other variables constant.
In a function \( f(x, y) \), the partial derivative \( \frac{\partial f}{\partial x} \) measures the rate of change of the function with respect to \( x \), and \( \frac{\partial f}{\partial y} \) measures it with respect to \( y \).

• To find these in our example function \( f(x, y) = A - (x^2 + Bx + y^2 + Cy) \), calculate: \[ \frac{\partial f}{\partial x} = -2x - B \quad \text{and} \quad \frac{\partial f}{\partial y} = -2y - C. \]
• Setting these to zero helps find the critical points like \( B = 4 \) and \( C = -2 \).
Partial derivatives allow us to explore how functions behave based on each dimension, facilitating the analysis of more complex systems.