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Partial derivatives are an essential tool in multivariate calculus that help to analyze functions with more than one variable. They allow us to determine how a function changes as each variable is varied, holding others constant. In this particular optimization problem, we are interested in finding the points on the cone closest to the point \((4, 2, 0)\) by minimizing the function \(D^2 = (x - 4)^2 + (y - 2)^2 + x^2 + y^2\).

• The partial derivative with respect to \(x\), represented as \(\frac{\partial D^2}{\partial x}\), tells us how \(D^2\) changes as \(x\) changes.
• Similarly, \(\frac{\partial D^2}{\partial y}\) indicates how \(D^2\) changes with respect to \(y\).

By setting these partial derivatives to zero, we find the critical points which are potential candidates for the minima or maxima of the function. In this problem, solving \(4x - 8 = 0\) leads to \(x = 2\), and from \(4y - 4 = 0\), we deduce \(y = 1\). These solutions represent the critical point which we further test to find the minimum distance.

Distance minimization is a common optimization problem that involves finding the point or points that minimize the distance to a given fixed point. In this scenario, we use the squared distance formula, $D^2 = (x - 4)^2 + (y - 2)^2 + z^2$, to simplify calculations, as it avoids the complexity of square roots.

• We aim to minimize the distance $D$, but by simplifying to $D^2$, we reduce the difficulty of differentiation.
• Begin by substituting $z^2 = x^2 + y^2$ from the cone equation.
• This substitution helps us rewrite the objective function entirely in terms of $x$ and $y$.

After substitution, we get $D^2 = x^2 - 8x + 16 + y^2 - 4y + 4 + x^2 + y^2$, which simplifies to $2x^2 + 2y^2 - 8x - 4y + 20$. This is the function we differentiate partially with respect to $x$ and $y$, and its critical points suggest where the minimum distance occurs.

The cone equation \(z^2 = x^2 + y^2\) defines the three-dimensional shape on which we must minimize the distance to a given point in space. This equation describes a right circular cone in a coordinate system where \(z\) is aligned with the cone's axis of symmetry.

• This equation is pivotal in transforming a three-variable problem into a simpler two-variable optimization task.
• By substituting \(z^2\) with \(x^2 + y^2\) in the distance formula, the complexity of solving the optimization problem reduces.
• Understanding how to manipulate such equations is crucial for solving real-world problems involving geometric constraints.

For this exercise, we ensure our solutions \((x, y, z)\) satisfy the cone equation after determining \(x\) and \(y\) from minimizing \(D^2\). With \(x = 2\) and \(y = 1\), \(z\) is computed by solving \(z^2 = 2^2 + 1^2 = 5\), giving us \(z = \pm\sqrt{5}\). These values satisfy the original cone equation, confirming they represent valid points on the cone.