Through market research, a company finds that it can expect to sell 45 – 5x products if each is priced at 1.25x dollars.
The correct answer and explanation is:
Let’s analyze the problem step-by-step.
Given:
- Demand function (quantity sold) = 45−5×45 – 5x
- Price per product = 1.25×1.25x dollars
Here, xx is some variable related to pricing (likely a price factor or a decision variable).
1. Expression for Revenue
Revenue RR is given by: R=(Price per product)×(Number of products sold)R = (\text{Price per product}) \times (\text{Number of products sold})
Substitute the given expressions: R=(1.25x)×(45−5x)=1.25x×(45−5x)R = (1.25x) \times (45 – 5x) = 1.25x \times (45 – 5x)
Simplify: R=1.25x×45−1.25x×5x=56.25x−6.25x2R = 1.25x \times 45 – 1.25x \times 5x = 56.25x – 6.25x^2
2. Interpretation
- xx affects both the price and the number of units sold.
- As xx increases, price per product increases (since price = 1.25×1.25x).
- However, the number sold decreases linearly by 5x5x.
- The revenue is a quadratic function in xx, opening downward (because the coefficient of x2x^2 is negative: −6.25-6.25), which means it has a maximum point.
3. Finding the Price and Quantity that Maximizes Revenue
To maximize revenue, differentiate R(x)R(x) with respect to xx and set to zero: dRdx=56.25−12.5x=0\frac{dR}{dx} = 56.25 – 12.5x = 0
Solve for xx: 12.5x=56.25 ⟹ x=56.2512.5=4.512.5x = 56.25 \implies x = \frac{56.25}{12.5} = 4.5
4. Calculate optimal price and quantity sold
- Price:
P=1.25×4.5=5.625 dollarsP = 1.25 \times 4.5 = 5.625 \text{ dollars}
- Quantity:
Q=45−5×4.5=45−22.5=22.5 unitsQ = 45 – 5 \times 4.5 = 45 – 22.5 = 22.5 \text{ units}
5. Maximum Revenue
Calculate revenue at x=4.5x = 4.5: R=56.25×4.5−6.25×(4.5)2=253.125−126.5625=126.5625 dollarsR = 56.25 \times 4.5 – 6.25 \times (4.5)^2 = 253.125 – 126.5625 = 126.5625 \text{ dollars}
Summary
- The company’s revenue as a function of xx is:
R=56.25x−6.25x2R = 56.25x – 6.25x^2
- Revenue is maximized when x=4.5x = 4.5.
- At this xx, the optimal price per product is $5.63 (rounded), and the expected sales volume is 22.5 units.
- The maximum revenue achievable is about $126.56.
Explanation (300 words):
The problem describes a company’s expected sales quantity and pricing strategy through a variable xx. The demand function is 45−5×45 – 5x, indicating that as xx increases, fewer products are sold — this models the typical negative relationship between price and quantity demanded. The price per product is 1.25×1.25x, which means the price increases proportionally with xx.
To find total revenue, multiply the number of products sold by the price per product, yielding a quadratic expression: R(x)=56.25x−6.25x2R(x) = 56.25x – 6.25x^2. This quadratic opens downward, indicating that revenue increases with xx up to a point and then decreases. To find the maximum revenue, we differentiate R(x)R(x) with respect to xx and set the derivative to zero. Solving this derivative gives the critical value x=4.5x=4.5.
Substituting x=4.5x=4.5 back into the price and quantity functions gives the price of approximately $5.63 and a sales volume of 22.5 units. This combination maximizes revenue, which is about $126.56.
This analysis helps the company understand the optimal pricing strategy to maximize revenue, balancing the trade-off between increasing prices and decreasing sales volume. Pricing too high will decrease quantity sold excessively, reducing revenue; pricing too low limits revenue per unit. This mathematical model guides strategic pricing decisions to optimize income.