Problem 22 - Contemporary Precalculus

Question

A manufacturer's revenue (in cents) from selling \(x\) items per week is given by \(200 x-.02 x^{2} .\) It costs \(60 x+30,000\) cents to make \(x\) items. (a) Approximately how many items should be made each week to make a profit of \(\$ 1100 ?\) (Don't confuse cents and dollars.) (b) How many items should be made each week to have the largest possible profit? What is that profit?

Step-by-Step Solution

Answer

Answer: Approximately 1,850 items need to be produced per week to have a profit of $1,100. The maximum possible weekly profit is $2,150, which can be achieved when producing 3,500 items per week.

Step 1: Find the profit function

To find the profit function, we need to subtract the cost function from the revenue function: Profit(x) = Revenue(x) - Cost(x) Remember to convert the given profit in part (a) from dollars to cents since the given functions are in cents. So, \( \$ 1100 = 1100\cdot 100 = 110,000\) cents.

Step 2: Determine the profit function

Now let's determine the profit function by substituting the given revenue and cost functions: Profit(x) = \((200x - 0.02x^2) - (60x + 30,000)\) Profit(x) = \(200x - 0.02x^2 - 60x - 30,000\) Profit(x) = \(140x - 0.02x^2 - 30,000\)

Step 3: Solve for x with a profit of 110,000 cents

To find the amount of items to be produced to have a profit of 110,000 cents, set the profit function equal to 110,000 cents and solve for x: \(110,000 = 140x - 0.02x^2 - 30,000\) \(0 = -0.02x^2 + 140x - 140,000\) To solve the quadratic equation for x, use either factoring, quadratic formula, or completing the square method. In this case, we'll use the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = -0.02, b = 140,\) and \(c = -140,000\)

Step 4: Calculate x using the quadratic formula

Substitute the values of a, b, and c into the quadratic formula and solve for x: \(x = \frac{-140 \pm \sqrt{140^2 - 4(-0.02)(-140,000)}}{2(-0.02)}\) \(x ≈ \frac{-140 \pm 600}{-0.04}\) There are two possible solutions for x: 1. \(x_1 ≈ \frac{-140 + 600}{-0.04} ≈ 11500\) 2. \(x_2 ≈ \frac{-140 - 600}{-0.04} ≈ 1850\) Choose the solution that makes sense within the context of the problem. In this case, \(x_2 ≈ 1850\) is the more reasonable number of items to be produced per week to give a profit of 110,000 cents or \( \$ 1100\).

Step 5: Find critical points of the profit function to maximize profit

To find the maximum profit, take the first derivative of the profit function and find the critical points: \(\frac{d(\text{Profit})}{dx} = \frac{d(140x - 0.02x^2 - 30,000)}{dx}\) \(\frac{d(\text{Profit})}{dx} = 140 - 0.04x\) Set the derivative equal to zero and solve for x: \(0 = 140 - 0.04x\) \(x = 3500\)

Step 6: Calculate the maximum profit

Substitute the value of x obtained from step 5 into the profit function to find the maximum profit: Maximum_Profit = Profit(3500) Maximum_Profit = \(140(3500) - 0.02(3500)^2 - 30,000\) Maximum_Profit = \(490,000 - 245,000 - 30,000\) Maximum_Profit = \(215,000\) cents To convert the maximum profit back to dollars, divide by 100: Maximum_Profit = \(2,150\) So, the largest possible profit is \(\$ 2,150\) per week when producing 3500 items per week.

Key Concepts

Revenue Function

The revenue function is a representation of the total income generated from selling a given number of items. It's crucial in understanding how sales activities translate to revenue. Here, the revenue function is given as \( R(x) = 200x - 0.02x^2 \), indicating that the revenue depends on both a linear component \(200x\) and a quadratic component \(-0.02x^2\). These components show how revenue increases with the number of items sold but at a decreasing rate due to the quadratic term.
This occurrence is typical when more items are produced and sold, leading to factors like market saturation or reduced prices that influence revenue.
Understanding the revenue model helps predict how changing the quantity sold impacts total revenue.

Cost Function

A cost function represents the total cost incurred in producing a certain amount of goods. In this exercise, the cost function is provided as \( C(x) = 60x + 30,000 \), where \(60x\) represents the variable costs dependent on the production of \(x\) items, and \(30,000\) is a fixed cost.

• **Variable Costs**: These are costs that change with the number of units produced, such as materials and labor.
• **Fixed Costs**: These remain constant irrespective of production levels, such as rent, salaries, etc.

Analyzing the cost function is critical for businesses to determine pricing strategies and production levels that not only cover these costs but also generate profit.

Quadratic Formula

The quadratic formula is a mathematical tool used to find the values of \(x\) that satisfy a quadratic equation of the form \(ax^2 + bx + c = 0\). The formula is:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]In the context of the exercise, it's applied to solve for \(x\) in the profit equation derived from: \(0 = -0.02x^2 + 140x - 140,000\)The use of the quadratic formula here helps find potential production quantities that meet specific profit targets. The roots of the equation, \(x_1\) and \(x_2\), help identify feasible solutions based on the scenario's context. It is vital to choose the solution that aligns with realistic constraints, such as production capacity and market demand.

Derivative

A derivative, in simple terms, helps us understand how a function changes at a particular point. In the context of profit maximization, the first derivative of the profit function is used to identify the rate of change of profit with respect to production volume. It assists in finding where this rate is zero (critical points), implying potential maximum or minimum profit points.
The derivative of \(P(x) = 140x - 0.02x^2 - 30,000\) is calculated as:\[ \frac{d(P)}{dx} = 140 - 0.04x \]Setting the derivative equal to zero helps find critical production values that maximize profit. Solving \(140 - 0.04x = 0\) allows businesses to determine optimal production levels.

Critical Points

Critical points are the values at which the first derivative of a function is zero. These are significant in profit maximization as they indicate where the profit is at a local maximum or minimum.
For the profit function \(P(x) = 140x - 0.02x^2 - 30,000\), the critical point is found by solving the equation derived from its first derivative:\[ 140 - 0.04x = 0 \]Which gives us \(x = 3500\) as a critical point.
At this production level, substituting back into the profit function gives the maximum potential profit. Recognizing critical points is important for operations strategy, ensuring resources are focused on production levels that achieve the highest return.