Question

The weekly cost \(C\) of producing \(x\) units in a manufacturing process is given by the function \(C(x)=70 x+800\) The number of units \(x\) produced in \(t\) hours is given by \(x(t)=40 t\) Find and interpret \((C \circ x)(t)\).

Short answer

\((C \circ x)(t) = 2800t + 800\). This means that the cost of production is $2800 per hour, with a fixed initial cost of $800.

Step-by-step solution

Write down the Cost Function and Units Produced Function

We have our cost function of producing \(x\) units as \(C(x) = 70x + 800\), and our function of units produced in \(t\) hours as \(x(t) = 40t\).

Substitute \(x(t)\) into \(C(x)\), Compute \((C \circ x)(t)\)

Substitute the function \(x(t)\) into the function \(C(x)\) to compute the composite function \((C \circ x)(t)\). This gives: \((C \circ x)(t) = C(x(t)) = 70(40t) + 800 = 2800t + 800\).

Interpret the Result

\((C \circ x)(t) = 2800t+800\) represents the cost of manufacturing per hour. Specifically, the cost of production increases by $2800 for every hour, with a fixed cost (or startup cost) of $800.

Key concepts

Cost Function

A cost function is a mathematical relationship that correlates the cost of production to the number of items produced. In our example, the cost function is given by the formula \(C(x)=70x+800\), where \(C(x)\) represents the total cost of producing \(x\) units. This function involves two components: a variable cost and a fixed cost. The variable cost, here \(70x\), depends on the quantity produced, representing the additional cost for each unit. The fixed cost, in this case \(800\), is the amount that does not change regardless of the units produced. It could include rent, utilities, or equipment that are necessary before any production begins.

Understanding the cost function is crucial for a business to estimate the expenses and to set the right pricing strategy. For instance, it helps in determining the break-even point, which is the number of units that must be sold to cover all costs.

Units Produced Function

In the context of our exercise, the units produced function models the number of units produced over time. Specifically, the function \(x(t)=40t\) indicates that t hours of labor results in the production of \(40t\) units. The '40' is the production rate per hour, meaning that for every additional hour of work, 40 more units are created. This function is also a linear one, suggesting that the production rate is constant over time.

Celar comprehension of this function assists in production planning and workforce allocation. By knowing how many units can be produced in a given timeframe, a company can schedule shifts, manage resources, and ensure timely delivery of products to customers.

Interpreting Functions

Interpreting functions is about making sense of the mathematical expressions and understanding their real-world implications. When we found the composite function \((C \circ x)(t) = 2800t + 800\), we're essentially combining the 'units produced' function and the 'cost function' to see how the total cost of production relates directly to time instead of units produced. The interpretation here is that \(2800t\) represents the total variable costs per hour, and \(800\) is the fixed cost.

This tells us that for every additional hour of production, the variable costs increase by \(2800, on top of the initial \)800 flat rate. By understanding this relationship, businesses can make informed decisions regarding how long to run production to meet demand without incurring unnecessary costs.

Algebraic Operations

Algebraic operations involve manipulating algebraic expressions using the standard mathematical operations: addition, subtraction, multiplication, and division. In this exercise, we performed a substitution, which is a fundamental algebraic operation, to find the composite function. We substituted the units produced function, \(x(t)\), into the cost function, \(C(x)\), which led us to \((C \circ x)(t) = 2800t + 800\).

• We multiplied \(70\) by \(40t\) to get \(2800t\) showing the cost increase per hour of production.
• We then added the fixed cost of \(800\) to complete our composite function.

These operations are at the heart of understanding and interpreting functions in algebra, as they enable us to connect different functions and see how they interact in practical situations.