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Chapter 3: Problem 11

Your neighbors operate a successful bake shop. One of their specialties is a very rich whipped-cream-covered cake. They buy the cakes from a supplier who charges \(\$ 6.00\) per cake, and they sell 200 cakes weekly at \(\$ 10.00\) each. Research shows that profit from the cake sales can be increased by increasing the price. Unfortunately, for every increase of \(\$ 0.50\) cents, sales will drop by seven cakes. a. What is the optimal retail price for a cake to obtain a maximum weekly profit? b. The supplier, unhappy with reduced sales, informs the owners that if they purchase fewer than 165 cakes weekly, the cost per cake will increase to \$7.50. Now what is the optimal retail price per cake, and what is the bake shop's total weekly profit? c. Situations like this occur regularly in retail trade. Discuss the implications of reduced sales with increased total profit versus greater sales with smaller profits. For example, a drop in the number of customers could mean fewer sales of associated products.

Short Answer

Expert verified
(a) Optimal price: $14 per cake. (b) Optimal price: $13.50 per cake, profit: $906 weekly. (c) Higher price results in fewer sales but increased profits; fewer customers may reduce additional sales.

Step by step solution

01

Establish the Profit Equation for (a)

The weekly profit can be calculated as the difference between sales revenue and cost. The profit equation in terms of price increase, \( x \), is: \[P(x) = (10 + 0.5x)(200 - 7x) - 6(200 - 7x)\] This equation factors in 200 cakes sold at \(\\(10\) each, with each \(\\)0.50\) increase decreasing sales by 7 cakes. The cost per cake is \(\$6\).
02

Simplify the Profit Equation for (a)

Expand the profit equation: \[P(x) = (2000 + 100x - 70x - 3.5x^2) - (1200 - 42x)\] This simplifies to: \[P(x) = 800 + 58x - 3.5x^2\] This is a quadratic equation in the form \(ax^2 + bx + c\).
03

Find the Maximum Profit for (a)

A quadratic in the form \(ax^2 + bx + c\) reaches its maximum at \(x = -\frac{b}{2a}\). Here, \(a = -3.5\) and \(b = 58\). Calculate \(x\): \[x = -\frac{58}{2(-3.5)} = \frac{58}{7} = 8.2857\] Round \(x\) to the nearest integer, \(x = 8\), meaning 8 increases.
04

Calculate Optimal Price for (a)

With 8 increments of \(\\(0.50\), the price increase is \(8 \times 0.50 = \\)4.00\), making the optimal price \(\\(10 + \\)4 = \$14\) per cake.
05

Establish the Profit Equation for (b)

If fewer than 165 cakes are ordered, the cost per cake rises to \$7.50. When \(7x\) is more than 35 (causing sales to drop below 165), substitute the cost: \[P(x) = (10 + 0.5x)(200 - 7x) - 7.5(200 - 7x)\]
06

Simplify and Solve for (b)

Simplify: \[P(x) = (2000 + 100x - 70x - 3.5x^2) - (1500 - 52.5x)\] \[P(x) = 500 + 52.5x - 3.5x^2\] Again, for max profit, solve \(x = -\frac{52.5}{2(-3.5)} = \frac{52.5}{7} = 7.5\). Round to \(x = 7\).
07

Calculate Results and Profit for (b)

The price increase is \(7 \times 0.50 = \\(3.50\); thus, the optimal price is \(\\)10 + \\(3.50 = \\)13.50\). Weekly sales are \(200 - 7 \times 7 = 151\) cakes. Calculate the profit: \[P(7) = 151 \times 13.5 - 151 \times 7.5 = \$906\]
08

Discuss the Implications for (c)

Reducing sales while increasing total profit can improve business stability by boosting per-unit earnings and reducing associated variable costs such as restocking and handling. However, fewer customers might lead to lower sales of complementary goods, affecting overall business vitality and customer retention.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Profit Equation
Understanding how profits are calculated is the key to maximizing them. In our problem, the profit equation calculates the weekly profit based on the number of cakes sold, the selling price, and the cost per cake.
The profit equation is:
  • Profit = Sales Revenue - Costs
For example, let's define the current scenario:
- Sales revenue comes from selling 200 cakes at $10 each.
  • This is multiplied by any increase in price, noted by the variable \(x\), due to the implemented pricing strategy.

    • The total cost is the number of cakes sold multiplied by the supplier cost per cake. By defining the profit equation as a function of the price increase, we can optimize it using algebraic techniques. Expanding and simplifying this equation helps us understand how each additional price bump affects total profit. Using variables and solving equations allow businesses to see potential profits from different pricing strategies.
    Quadratic Equation
    Quadratic equations are pivotal in scenarios like determining maximum profit. Quadratic equations have a parabolic curve, and their peak can either be a maximum or minimum depending on the nature of the graph.
    In this context, the profit equation eventually simplifies into a quadratic form:
    • \(P(x) = ax^2 + bx + c\)
    Here, the coefficients \(a\), \(b\), and \(c\) represent specific aspects of the problem (like price increase impact and costs).Using the vertex formula \(x = -\frac{b}{2a}\), we find where the curve of the quadratic graph reaches its highest point.
    Understanding this allows us to discover the exact number of price increases that will bring about maximum profit. Real-world applications of this are numerous, especially in retail and other domains where pricing adjustments significantly affect revenues.
    Retail Pricing Strategy
    Strategizing the retail price of products like cakes involves more than just cranking up the numbers. It requires analyzing multiple factors that influence the sales and costs.
    In our scenario, raising the cake price by $0.50 reduces sales by 7 cakes per increment.
    • This trade-off needs careful consideration since it directly impacts both the revenue and customer buying behavior.
    Retail pricing strategies target not just immediate profit but long-term business stability.
    For instance, selling fewer cakes with a higher profit margin per cake might enhance financial health compared to selling more cakes with minimal profit. Such strategies can also affect associated sales; fewer customers might mean reduced sales of other in-store items. Thus, a strategic pricing approach optimally balances increased per-unit profit with maintaining customer flow, which is critical for sustaining business growth.

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    Most popular questions from this chapter

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