Problem 47 - Essential Calculus Early Transcendentals

Question

A manufacturer has been selling 1000 flat-screen TVs a weck at \(\$ 450\) cach. A market survey indicates that for cach \(\$ 10\) rebate offered to the buyer, the number of TVs sold will increase by 100 per week. (a) Find the demand function. (b) How large a rebate should the company offer the buyer in order to maximize its revenue? (c) If its weekly cost function is \(C(x)=68,000+150 x\) how should the manufacturer set the size of the rebate in order to maximize its profit?

Step-by-Step Solution

Answer

(a) Demand function: \(x = 5500 - 10p\). (b) Rebate of \$175 maximizes revenue. (c) Rebate of \$100 maximizes profit.

Step 1: Express Variables and Understand Relationships

Let the number of TVs sold per week be \(x\). Initially, \(x = 1000\) when price per TV is \(p = 450\). The TV's price decreases by \(10d\), where \(d\) is the number of \$10 rebates, and the number of TVs sold increases by \(100d\). Hence, \(x = 1000 + 100d\) and \(p = 450 - 10d\).

Step 2: Find the Demand Function

The demand function can be expressed as \(x = f(p)\), where \(x = 1000 + 100d\) and \(p = 450 - 10d\). We can solve for \(d\) as a function of \(p\) from \(p = 450 - 10d\): \(d = \frac{450 - p}{10}\). Substitute this into \(x = 1000 + 100d\): \(x = 1000 + 100\left(\frac{450 - p}{10}\right) = 1000 + 10(450 - p)\). Therefore, \(x = 5500 - 10p\).

Step 3: Determine Revenue Function

Revenue \(R\) is given by price multiplied by quantity, \(R(x) = x \cdot p\). From \(x = 5500 - 10p\), we solve for \(p\): \(p = \frac{5500 - x}{10}\). Thus, \(R(x) = x \cdot \left(\frac{5500 - x}{10}\right) = 550x - 0.1x^2\).

Step 4: Maximize Revenue

To maximize revenue, we take the derivative of \(R(x)\) and set it to zero: \(R'(x) = 550 - 0.2x = 0\). Solving for \(x\), \(x = 2750\). The number of TVs are maximized when \(x=2750\). To find the corresponding rebate, solve \(2750 = 1000 + 100d\), giving \(d = 17.5\). Therefore, the rebate is \(\$175\).

Step 5: Derive Profit Function

Profit \(P(x)\) is revenue minus cost: \(P(x) = R(x) - C(x) = (550x - 0.1x^2) - (68000 + 150x)\). Simplifying gives \(P(x) = 400x - 0.1x^2 - 68000\).

Step 6: Maximize Profit

To maximize profit, take the derivative of profit and set to zero: \(P'(x) = 400 - 0.2x = 0\). Solve \(P'(x) = 0\) gives \(x = 2000\). Solve for corresponding \(d\) from \(2000 = 1000 + 100d\) yields \(d = 10\). Thus, the rebate should be \(\$100\).

Key Concepts

Demand Function

The demand function is a mathematical relationship that illustrates how the quantity demanded of a good is influenced by its price. In simpler terms, it tells us how many units of a product consumers are willing to buy at different price points. Understanding the demand function is crucial for businesses as it helps them predict sales and make informed pricing decisions.

In the given exercise, we first establish the initial conditions: 1000 TVs are sold at \(450 each weekly. If a rebate of \)10 is offered, the sales increase by 100 units. Our task is to express the number of TVs sold, denoted by \(x\), as a function of the price \(p\). Starting with the relationships \(x = 1000 + 100d\) and \(p = 450 - 10d\), we derive \(d\) in terms of \(p\):
\[d = \frac{450 - p}{10}\]
This is substituted into the equation for \(x\), resulting in the demand function:
\[x = 5500 - 10p\]
This equation reflects how sales quantity increases as prices decrease. By interpreting the demand function, we can visualize the trade-offs between price and the quantity demanded.

Cost Function

The cost function represents the total cost incurred by a company to produce a particular number of goods. It includes fixed costs, which are costs that do not change with the level of output, and variable costs, which do increase as production increases.

In our problem, the cost function is given as:
\[C(x) = 68000 + 150x\]
Here, 68,000 is the fixed cost that the company must pay regardless of how many TVs they produce. The term \(150x\) reflects the variable cost, where 150 is the cost to produce each additional TV. Therefore, if the manufacturer increases the number of TVs sold, the production costs will also rise accordingly.

Understanding cost functions helps businesses in budgeting and setting prices to ensure they at least cover production costs and hopefully, make a profit. Efficient management can reduce these costs and maximize profitability.

Profit Maximization

Profit maximization is the process of identifying the price and production level that produces the highest profit for a firm. It’s a fundamental goal for any business, as profit represents the earnings left over after all costs have been paid.

For our exercise, the profit function \(P(x)\) is derived by subtracting the cost function from the revenue function. The expression becomes:
\[P(x) = 400x - 0.1x^2 - 68000\]
To find the profit-maximizing output level \(x\), we calculate the derivative of the profit function and solve for when it equals zero:
\[P'(x) = 400 - 0.2x = 0\]
Solving gives \(x = 2000\). This indicates that the manufacturer should sell 2000 TVs to maximize profit, and to achieve this sales volume, a rebate of \(\$100\) should be offered, derived from \(x = 1000 + 100d\) when \(x = 2000\).

This understanding allows businesses to adjust their strategies and optimize their operations for maximum financial benefit.