91Ó°ÊÓ

Calculus is a branch of mathematics that deals with change and motion. It is divided into two parts: differential calculus and integral calculus. In solving optimization problems, like finding the dimensions for the storage shed, we primarily focus on differential calculus. Calculus provides tools to calculate derivatives, which show how a function changes as its input changes. When we talk about optimization in calculus, it usually involves finding the maximum or minimum values of a function. These are found by identifying critical points where the derivative equals zero or doesn't exist. This is crucial in economic and physical applications, where we need to maximize efficiency or minimize cost, as in our storage shed example.

In the context of our storage shed problem, a volume constraint ensures the shed's volume is exactly 900 cubic feet. A constraint is something that must remain true while we adjust other variables, which in this case, are the shed's dimensions. To express the volume constraint, we start with the formula for the volume of a rectangular box:

• Volume = Length \( \times \) Width \( \times \) Height

Substituting the known values, we introduce the relationships between the dimensions. For our case:

• Width \( W = \frac{3}{4} L \) (since width is three-fourths the length)
• And given volume: \( L \times W \times H = 900 \)

This equation helps us express one dimension in terms of the others, assisting in constructing the cost function which we'll optimize.

A cost function helps us quantify the expense of building the storage shed, based on material costs per square foot for various parts: floor, walls, and roof. The goal of an optimization problem like ours is to find dimensions that minimize this function. The cost function can be set up by calculating the total expenses:

• Floor cost: based on area \( L \times W \), costing \( 4 \cdot \frac{3}{4}L^2 \)
• Wall costs: the sum of the two pairs of opposite walls. Each costs \( L \times H \) and \( W \times H \), with material cost \( 6 \) per square foot
• Roof cost: also based on area \( L \times W \), costing \( 3 \cdot \frac{3}{4}L^2 \)

Minimizing the total calculated cost will give us the most resource-efficient shed dimensions.

Differentiation is a fundamental part of calculus used to determine how a function changes as its input variable changes. In our optimization problem, we differentiate the cost function with respect to the length \( L \) of the shed, which helps identify critical points where the cost function could reach a minimum:

• Find the derivative: \( \frac{dC}{dL} \)
• Set the derivative equal to zero to find critical points: \( \frac{21}{2}L - \frac{12600}{L^2} = 0 \)

By solving this derivative equation, we can find the value of \( L \) which minimizes the cost. This process effectively explores how slight changes in \( L \) impact the overall cost, steering us towards an optimal solution.

The second derivative test is a method used after finding a critical point to determine if it is a local maximum or minimum. Once the first derivative is set to zero, we apply this second derivative test to ensure the critical point minimizes the cost function for constructing the shed:

• Calculate the second derivative: \( \frac{d^2C}{dL^2} \)
• Evaluate it at the critical point.

For this problem, the second derivative comes out to be:\[ \frac{21}{2} + \frac{25200}{L^3} \]Evaluate at our critical \( L \approx 10.46 \), which must be greater than zero. This positive value confirms that our critical point is indeed a local minimum, suggesting the dimensions found will minimize the cost effectively.