Problem 49 - Calculus of a Single Variable

Question

Minimum Distance In Exercises \(49-51\) , consider a fuel distribution center located at the origin of the rectangular coordinate system (units in miles; see figures). The center supplies three factories with coordinates \((4,1),(5,6),\) and \((10,3) .\) A trunk line will run from the distribution center along the line \(y=m x\) , and feeder lines will run to the three factories. The objective is to find \(m\) such that the lengths of the feeder lines are minimized. Minimize the sum of the squares of the lengths of the vertical feeder lines (see figure) given by $$S_{1}=(4 m-1)^{2}+(5 m-6)^{2}+(10 m-3)^{2}$$ Find the equation of the trunk line by this method and then determine the sum of the lengths of the feeder lines.

Step-by-Step Solution

Answer

The equation of the line is \(y = ((26 - 4sqrt{51})/19)x\). The sum of the lengths of the feeder lines is \((4*(26 - 4sqrt{51}))/19 - 1 + (5*(26 - 4sqrt{51})/19 - 6) + (10*(26 - 4sqrt{51})/19 - 3)\).

Step 1: Differentiate S1

Differentiate the formula for \(S_1\) with respect to \(m\). This gives \(S'_1 = 2(4m - 1) * 4 + 2(5m - 6) * 5 + 2(10m - 3) * 10\).

Step 2: Set the derivative to zero

To find the values of \(m\) for which \(S_1\) is minimum, set \(S'_1 = 0\) and solve for \(m\), which gives \(m = (26 - 4sqrt{51})/19\).

Step 3: Determine the equation of the line

Because \(m\) represents the slope of the line, the equation of the line is simply \(y = mx\). Substituting the value of \(m\) from step 2 gives us the equation \(y = ((26 - 4sqrt{51})/19)x\).

Step 4: Determine the sum of the lengths of the feeder lines

With the values of \(m\) and \(x\) known, we can determine the lengths of the feeder lines, which are the distances from the y-values (1, 6, 3) to the line's corresponding y-values at \(x = 4, 5, 10\). This gives us \((4*(26 - 4sqrt{51}))/19 - 1\), \((5*(26 - 4sqrt{51})/19 - 6\), and \((10*(26 - 4sqrt{51})/19 - 3)\). Adding these three values gives the sum of the lengths of the feeder lines.

Key Concepts

Derivative Optimization

Understanding how to optimize a function via its derivative is crucial in various mathematical and real-life applications. In the context of the given problem, we are looking to minimize the sum of the squares of the lengths of the vertical feeder lines.
To achieve this, we employ derivative optimization, which is the process of finding the minimum or maximum of a function by taking its derivative and setting it to zero. The function here represents the sum of the squares of the lengths, denoted as \(S_1\). The first step involves differentiating \(S_1\) with respect to \(m\), which gives us the rate at which \(S_1\) changes with changes in \(m\).
When we set the derivative, \(S'_1\), to zero and solve for \(m\), we are finding the slope value that will result in the feeder lines' lengths being the shortest. This is how we use the derivative—by identifying the 'critical points', which can be potential minima or maxima. In cases where there are multiple critical points, further analysis, like the second derivative test, can confirm whether the critical points are minima or maxima. For the given exercise, setting \(S'_1 = 0\) and solving for \(m\) helps us find the specific slope that minimizes the feeder lines' lengths.

Slope of a Line

The slope of a line is one of the most fundamental concepts in algebra and geometry, representing the steepness and direction of the line. Geometrically, it's the ratio of the rise over the run between any two points on the line. In equation form, it's expressed as \(m\) in the linear equation \(y = mx + b\), where \(b\) is the y-intercept of the line.
In our problem, we seek the optimal value of \(m\) for the trunk line that will minimize the total length of the feeder lines. The trunk line's equation is given by \(y = mx\), indicating that it passes through the origin, as there is no \(b\) term (i.e., the y-intercept is zero). After optimizing \(m\) through the derivative process, we plug this value into the trunk line's equation to get the final equation of the line. Then, by using the slope-intercept form, we assess how the line relates to the feeder lines drawn to the factories, thus informing the length and configuration of those feeder lines.

Minimizing Sum of Squares

Minimization problems often involve minimizing the sum of squares since squares can aid in measuring the 'distance' from a desired value or condition. For instance, in our problem, the square of a feeder line's length corresponds to the squared difference between the factory's y-coordinate and the trunk line's y-value at the respective x-coordinate of the factory.
By squaring these distances, we are creating a function, \(S_1\), that when minimized, reduces the variation between our data points (the factories' y-coordinates) and the model (the trunk line). This approach is widely applicable, such as in the method of least squares, commonly used in statistical regression analysis.
To minimize \(S_1\), we use calculus to find the derivative of \(S_1\) with respect to \(m\) and determine the slope that results in the least total squared distance. This technique effectively 'fits' our trunk line in the best possible manner relative to the fixed factory locations, thereby minimizing the sum of the lengths of the feeder lines connecting to it.