Question

Maximizing a multivariate function without constraints. Find the maximum \(\left(x^{*}, y^{*}, z^{*}\right)\) of the function \(f(x, y, z)=d-(x-a)^{2}-(y-b)^{2}-(z-c)^{2} .\)

Short answer

The maximum is at \((a, b, c)\) with the value \(f(a, b, c) = d\).

Step-by-step solution

- Identify the function and given variables

The function to maximize is \(f(x, y, z) = d - (x - a)^2 - (y - b)^2 - (z - c)^2\). The variables are \(x, y, z\), and the constants are \(a, b, c, d\).

- Find the first partial derivatives

Calculate the first partial derivatives of \(f(x, y, z)\) with respect to \(x\), \(y\), and \(z\).\( \frac{\partial f}{\partial x} = -2(x - a) \)\( \frac{\partial f}{\partial y} = -2(y - b) \)\( \frac{\partial f}{\partial z} = -2(z - c) \)

- Set partial derivatives equal to zero

Set each of the first partial derivatives equal to zero to find the critical points.\( -2(x - a) = 0 \)\( -2(y - b) = 0 \)\( -2(z - c) = 0 \)

- Solve for \(x, y, z\)

Solve the equations from Step 3 for \(x, y, z\).\( x - a = 0 \implies x = a \)\( y - b = 0 \implies y = b \)\( z - c = 0 \implies z = c \)

- Verify the solution

Since the second partial derivatives are all negative, the function \(f(x, y, z)\) is concave and the critical point \((a, b, c)\) is a maximum. The maximum value is \(f(a, b, c) = d\).

Key concepts

Partial Derivatives

In multivariate calculus, partial derivatives represent the rate at which a function changes as one of its variables is varied, while other variables are kept constant. For the given function, we calculated the partial derivatives with respect to each variable.

For example, the partial derivative of the function with respect to x is \(\frac{\text{∂}f}{\text{∂}x} = -2(x - a)\).
Here, we treat y and z as constants. This derivative tells us how the function changes as x changes, isolating the effect of x from y and z. We did similarly for y and z.

Calculating these derivatives is essential because the points where these partial derivatives equal zero signal where the function might have extremum points like minima or maxima.

Critical Points

Critical points are specific points in the domain of a function where its partial derivatives are zero. These points are essential for determining potential maxima, minima, or saddle points.

In our exercise, setting each first partial derivative to zero led us to the system of equations:

• \(-2(x - a) = 0\)
• \(-2(y - b) = 0\)
• \(-2(z - c) = 0\)

Solving these, we found the critical point at
\

Concavity

Concavity helps determine whether a critical point is a maximum, minimum, or saddle point by examining the second partial derivatives. For the given function, we need to check the outcome of the second partial derivatives.

Here, the second partial derivatives of the function are:

• \( \frac{\text{∂}^{2}f}{\text{∂}x^{2}} = -2 \)
• \( \frac{\text{∂}^{2}f}{\text{∂}y^{2}} = -2 \)
• \( \frac{\text{∂}^{2}f}{\text{∂}z^{2}} = -2 \)

Since these are all negative, the function is concave downwards. This fact implies that the critical point \( (a, b, c) \) is indeed a maximum.
Hence, verifying the solution shows that \(f(a, b, c) \) achieves its maximum value, which is \d\.