Problem 46 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

Oil from an offshore rig located 3 miles from the shore is to be pumped to a location on the edge of the shore that is 8 miles east of the rig. The cost of constructing a pipe in the ocean from the rig to the shore is \(1.5\) times more expensive than the cost of construction on land. How should the pipe be laid to minimize cost?

Step-by-Step Solution

Answer

The pipe should be laid 2.68 miles east from the point directly east of the rig on the shore.

Step 1 - Define Variables

Let the point directly east from the rig on the shore be point A, and let the point where the pipe hits the shore be point B, which is x miles east of point A. The total distance east on the shore to the point where the pipe hits the shore is 8 miles.

Step 2 - Ocean and Land Sections

The pipe will be laid in two sections: from the rig to point B in the ocean, and from point B to the delivery point on land. Calculate the distance from the rig to point B using Pythagoras’ theorem as \(\sqrt{x^2+9}\) miles (considering the vertical distance from the rig to the shore is 3 miles).

Step 3 - Cost Function Setup

The cost function will take into account the different costs of laying pipe in the ocean and on land. Letting the cost per mile on land be C, the cost per mile in the ocean will be 1.5C. Therefore, the total cost function is \(C_{\text{total}}(x) = 1.5C\sqrt{x^2+9} + C(8-x)\).

Step 4 - Simplify the Cost Function

Simplify the cost function: \(C_{\text{total}}(x) = 1.5C\sqrt{x^2+9} + 8C - Cx\).

Step 5 - Find Derivative of Cost Function

To minimize the cost, find the critical points by taking the derivative of the cost function with respect to x. Use the chain rule for the square root term: \(\frac{d}{dx}C_{\text{total}}(x) = \frac{d}{dx}(1.5C\sqrt{x^2+9}) - C\).

Step 6 - Solve for Critical Points

Set the derivative equal to zero to find the critical points. \(1.5C \frac{2x}{2\sqrt{x^2+9}} - C = 0\). Simplify to \(\frac{1.5Cx}{\sqrt{x^2+9}} = C\). So, \(1.5x = \sqrt{x^2+9}\).

Step 7 - Solve the Equation

Square both sides to remove the square root: \(2.25x^2 = x^2 + 9\). Simplify to get \(1.25x^2 = 9\) and solve for x: \(x^2 = \frac{9}{1.25} = 7.2\). Therefore, x = \(\sqrt{7.2} \approx 2.68\).

Step 8 - Verify and Interpret

Verify that the critical point corresponds to a minimum by checking the second derivative or examining the behavior of the first derivative around the critical point. Since laying the pipe further east or further west leads to higher costs, x = 2.68 miles east of point A is optimal.

Key Concepts

Pythagorean Theorem

The Pythagorean Theorem is essential in this optimization problem. It helps us find the straight-line distance (hypotenuse) from the offshore rig to the point where the pipe hits the shore.
We know the direct distance from the rig to the shore is 3 miles, and the horizontal distance from point A to B on the shore is x miles. We use the Pythagorean Theorem as follows:

• Let the distances be the sides of a right triangle.
• The hypotenuse (distance from the rig to point B) is calculated as \(\sqrt{x^2 + 9}\).

This calculation is critical when setting up the cost function.

Cost Function

In this problem, we need to minimize the cost of laying the pipe. Here's how we set up a cost function that takes into account different costs for ocean and land sections.

• Let the cost per mile on land be C.
• The cost per mile in the ocean is 1.5C.

The total cost function thus becomes:
\[ C_{\text{total}}(x) = 1.5C \sqrt{x^2 + 9} + C (8 - x) \]
This function considers the distance of the pipeline both in the ocean and on land, multiplied by their respective costs.

Derivative

To minimize the construction cost, we need to find the critical points. This involves taking the derivative of the cost function with respect to x. The derivative of the cost function is calculated using the chain rule for the square root term:
\[ \frac{d}{dx} C_{\text{total}}(x) = \frac{d}{dx}(1.5C \sqrt{x^2 + 9}) - C \]
Using the chain rule, we break the differentiation into two parts:

• Exterior function: Differentiating the square root term, \( \sqrt{x^2 + 9} \)
• Interior function: Differentiating the inner term \( x^2 + 9 \)

Applying the chain rule, we get:
\[ 1.5C \frac{2x}{2\sqrt{x^2 + 9}} - C \]

Critical Points

Critical points are where the derivative equals zero. To find these points, set the derivative equal to zero:
\[ 1.5C \frac{x}{\sqrt{x^2 + 9}} - C = 0 \]
Simplify this equation to find:\
\[ \frac{1.5Cx}{\sqrt{x^2 + 9}} = C \]
Solving for x, we get:
\[ 1.5x = \sqrt{x^2 + 9} \]
Square both sides to clear the square root:
\[ 2.25x^2 = x^2 + 9 \]
Solving, we find:
\[ x^2 = 7.2 \]
So, \( x = \sqrt{7.2} \approx 2.68 \) miles east of point A.

Chain Rule

The chain rule is a fundamental technique in calculus used to differentiate composite functions. In this context, the chain rule helps us take the derivative of the square root term in the cost function.
Consider the function \( h(x) = 1.5C \sqrt{x^2 + 9} \). We need to differentiate it with respect to x.
The original function is a composition of:

• An outer function: \( f(u) = \sqrt{u} \)
• An inner function: \( g(x) = x^2 + 9 \)

The chain rule states:
\[ h'(x) = f'(g(x)) \cdot g'(x) \]
For our function:
\[ \frac{d}{dx}(1.5C \sqrt{x^2 + 9}) = \frac{1.5C \cdot 0.5 \cdot (2x)}{\sqrt{x^2 + 9}} = \frac{1.5C \cdot x}{\sqrt{x^2 + 9}} \]