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Difficulty: HardAlgebraic Word Problems and Modeling

An electronics manufacturer models the monthly demand NN for a specific model of tablet as N(P)=1,200−3PN(P) = 1,200 - 3P, where PP is the selling price per tablet in dollars. The monthly total cost of producing these tablets consists of a fixed overhead cost of $15,000\$15,000 plus a variable cost of $100\$100 per tablet produced. If the manufacturer earned a monthly net profit of $45,000\$45,000, what is the smaller of the two possible selling prices, in dollars, that could yield this profit?

Answer: 200 dollars

Answer

The smaller of the two possible selling prices is 200 dollars.
Total revenue is given by R(P)=P⋅N(P)=P(1,200−3P)=1,200P−3P2R(P) = P \cdot N(P) = P(1,200 - 3P) = 1,200P - 3P^2, and total cost is given by C(P)=15,000+100(1,200−3P)=135,000−300PC(P) = 15,000 + 100(1,200 - 3P) = 135,000 - 300P. The profit equation Profit(P)=R(P)−C(P)=45,000\text{Profit}(P) = R(P) - C(P) = 45,000 yields −3P2+1,500P−135,000=45,000-3P^2 + 1,500P - 135,000 = 45,000. Rearranging gives −3P2+1,500P−180,000=0-3P^2 + 1,500P - 180,000 = 0, which simplifies to P2−500P+60,000=0P^2 - 500P + 60,000 = 0. Factoring gives (P−200)(P−300)=0(P - 200)(P - 300) = 0, leading to prices of $200\$200 and $300\$300. The smaller price is $200\$200.

Step-by-Step Solution

1
Formulate algebraic functions for monthly revenue and monthly total cost in terms of price PP.
Revenue R(P)=P(1,200−3P)=1,200P−3P2R(P) = P(1,200 - 3P) = 1,200P - 3P^2 and Total Cost C(P)=15,000+100(1,200−3P)=135,000−300PC(P) = 15,000 + 100(1,200 - 3P) = 135,000 - 300P.
Total revenue equals unit price multiplied by demand quantity, while total cost equals fixed overhead plus variable cost per unit multiplied by demand quantity.
2
Formulate the net profit function Profit(P)=R(P)−C(P)\text{Profit}(P) = R(P) - C(P) and equate it to the target profit of $45,000\$45,000.
(1,200P - 3P^2) - (135,000 - 300P) = 45,000 ,whichsimplifiesto, which simplifies to -3P^2 + 1,500P - 180,000 = 0$.
Net profit is defined as total revenue minus total cost.
3
Divide the resulting quadratic equation by −3-3 to normalize coefficients and solve for PP.
P^2 - 500P + 60,000 = 0 \implies (P - 200)(P - 300) = 0 ,givingroots, giving roots P = 200 and and P = 300$.
Simplifying the quadratic equation reveals two factored linear roots.
4
Identify the smaller of the two valid price solutions.
P = 200 dollars.
The question specifically requests the smaller of the two possible selling prices.

Key Concept

Algebraic Modeling of Revenue, Cost, and Profit Functions using Quadratic Equations
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Topics

Algebraic Word Problems and Modeling

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