Problem 18 - Applied Calculus

Question

A missile has a guidance device which is sensitive to both temperature, \(t^{\circ} \mathrm{C},\) and humidity, \(h .\) The range in \(\mathrm{km}\) over which the missile can be controlled is given by Range \(=27,800-5 t^{2}-6 h t-3 h^{2}+400 t+300 h\) What are the optimal atmospheric conditions for controlling the missile?

Step-by-Step Solution

Answer

Optimal conditions are \(t = 40^{\circ}C\) and \(h = 10\).

Step 1: Understanding the Problem

We need to find the optimal atmospheric conditions, specifically the values for temperature \(t\) and humidity \(h\), that maximize the missile's control range. This requires finding the maximum value of a quadratic function in two variables.

Step 2: Express Range Equation

The range equation is given as:\[R(t, h) = 27,800 - 5t^2 - 6ht - 3h^2 + 400t + 300h.\] Our goal is to maximize \(R\) with respect to both \(t\) and \(h\).

Step 3: Partial Derivatives Calculation

To find the conditions for maximum range, calculate the partial derivatives of \(R\) with respect to \(t\) and \(h\). Set them equal to zero to find the critical points. \(\frac{\partial R}{\partial t} = -10t - 6h + 400 = 0\)\(\frac{\partial R}{\partial h} = -6t - 6h + 300 = 0\)

Step 4: Solving the System of Equations

Solve the system of equations:1. \(-10t - 6h + 400 = 0\)2. \(-6t - 6h + 300 = 0\)From equation (2):\(-6t - 6h + 300 = 0 \Rightarrow t + h = 50\)Substitute into equation (1):\(-10t - 6(50 - t) + 400 = 0\), leading to \(t = 40\) and \(h = 10\).

Step 5: Confirm Maximum with Second Derivative Test

Check the second derivatives:\(\frac{\partial^2 R}{\partial t^2} = -10\), \(\frac{\partial^2 R}{\partial h^2} = -6\), and mixed partial derivative \(\frac{\partial^2 R}{\partial t\partial h} = -6\). The Hessian determinant is positive, confirming a local maximum.

Key Concepts

Quadratic Functions

Quadratic functions are mathematical expressions where the highest degree of any variable is two, generally given by the form \( ax^2 + bx + c \). In our missile range problem, the function is slightly more complex due to its variables, namely temperature \(t\) and humidity \(h\), both squared:
\[ R(t, h) = 27,800 - 5t^2 - 6ht - 3h^2 + 400t + 300h \]
This expression involves not only separate squares of \(t\) and \(h\), but also a term that combines both variables, the \( -6ht \) term. Such functions are called quadratic in two variables.
To maximize this range, one must consider how both \(t\) and \(h\) simultaneously impact the outcome. This requires more advanced calculus tools compared to functions of a single variable. Our task is to find a 'maximum' point where this quadratic function achieves its highest value under given constraints, which involves understanding and applying calculus concepts like derivatives.

Partial Derivatives

Partial derivatives are used when a function has several variables, and you want to understand how the function changes as only one of these variables changes, keeping all others constant.
In the missile problem, the function \( R(t, h) \) involves two variables. For optimization, we find its partial derivatives with respect to \(t\) and \(h\), noted as \( \frac{\partial R}{\partial t} \) and \( \frac{\partial R}{\partial h} \).
This process involves treating all other variables as constants while differentiating:

• \( \frac{\partial R}{\partial t} = -10t - 6h + 400 \)
• \( \frac{\partial R}{\partial h} = -6t - 6h + 300 \)

Setting these derivatives equal to zero allows us to find critical points, potential maxima or minima of the function. Solving these equations provides the conditions where the control range is optimized — in this case, solving tells us \(t = 40\) and \(h = 10\) maximize the range, indicating optimal temperature and humidity values.

Hessian Determinant

Once we have found the critical points using partial derivatives, confirming whether these points are maxima, minima, or saddle points involves the Hessian determinant. The Hessian matrix is a square matrix of second-order partial derivatives. For a function \( f(x, y) \), the Hessian looks like this:
\[\begin{bmatrix} \frac{\partial^2 f}{\partial x^2} & \frac{\partial^2 f}{\partial x \partial y} \ \frac{\partial^2 f}{\partial y \partial x} & \frac{\partial^2 f}{\partial y^2}\end{bmatrix}\]
In our case with \( R(t, h) \), the second derivatives are:

• \( \frac{\partial^2 R}{\partial t^2} = -10 \)
• \( \frac{\partial^2 R}{\partial h^2} = -6 \)
• \( \frac{\partial^2 R}{\partial t \partial h} = -6 \)

The Hessian determinant is calculated as:
\[ D = \frac{\partial^2 R}{\partial t^2} \cdot \frac{\partial^2 R}{\partial h^2} - \left(\frac{\partial^2 R}{\partial t \partial h}\right)^2 \]
Substituting the values, we find that \( D = (-10)(-6) - (-6)^2 = 60 - 36 = 24 \).
Since \(D > 0\) and the leading diagonal term \( \frac{\partial^2 R}{\partial t^2} \) is negative, this confirms a local maximum at our critical point \(t = 40\) and \(h = 10\). Thus, the conditions are optimal for the missile's range.