Profit as a parabola

Noa prints T-shirts and sells them online.
At a price of $\displaystyle p$ shekels a shirt she sells $\displaystyle {q}={200-4p}$ shirts a week.
Each shirt costs her 15 ₪ (the shirt and the ink).
Her fixed costs – hiring the machine and the website – are 600 ₪ a week.

• The cost is $\displaystyle 600+15q$: the fixed costs and 15 ₪ for every shirt.
• The profit is the revenue minus the cost.
• The greatest revenue in the table is at 25 ₪, but the greatest profit is between 30 ₪ and 35 ₪.

$\displaystyle R$ – the revenue, $\displaystyle C$ – the cost. A negative profit is a loss.

We write all three as functions of the price $\displaystyle p$:

• The revenue: $\displaystyle {R(p)}={p(200-4p)}={200p-4p^{2}}$.
• The cost: $\displaystyle {C(p)}={600+15(200-4p)}={600+3000-60p}={3600-60p}$.
• The profit: $\displaystyle {P(p)}={200p-4p^{2}-(3600-60p)}={-4p^{2}+260p-3600}$.

The cost falls as the price rises, because fewer shirts sell. The profit is a parabola opening downwards, like the revenue.

On the graph, $\displaystyle x$ is the price of a shirt, and $\displaystyle y$ the shekels a week: the profit $\displaystyle P$, and the revenue $\displaystyle R$ (the dashed one).

• $\displaystyle V$ is the vertex of the profit’s parabola: at a price of 32.5 ₪ the profit is 625 ₪ a week, the greatest.
• $\displaystyle A$ and $\displaystyle B$ are where the profit’s parabola meets the $\displaystyle x$-axis: at 20 ₪ and at 45 ₪ the profit is 0.
• The revenue’s vertex, at 25 ₪, is to the left of the profit’s.

The price that gives the greatest profit is the $\displaystyle x$-coordinate of the vertex of $\displaystyle P$:

• At this price $\displaystyle {200-4\cdot32.5}={70}$ shirts sell a week.
• The profit: $\displaystyle {P(32.5)}={-4\cdot32.5^{2}+260\cdot32.5-3600}={625}$ ₪.
• The most profitable price is higher than the price that gives the greatest revenue (25 ₪): every shirt sold costs 15 ₪, so it pays to sell fewer shirts at a higher price.

At a break-even price the revenue equals the cost, and the profit is 0. These are the solutions of the equation $\displaystyle {P(p)}={0}$:

• We divide by $\displaystyle (-4)$: $\displaystyle {p^{2}-65p+900}={0}$.
• By the quadratic formula: $\displaystyle {p}={\frac{65\pm\sqrt{4225-3600}}{2}}={\frac{65\pm25}{2}}$.
• $\displaystyle {p}={20}$ or $\displaystyle {p}={45}$.

At any price between 20 ₪ and 45 ₪ Noa makes a profit. Below 20 ₪ the revenue does not cover the cost, and above 45 ₪ too few shirts sell – either way there is a loss.

A cake shop sells $\displaystyle {q}={100-2p}$ cakes a day, where $\displaystyle p$ is the price of a cake in shekels.
Each cake costs it 20 ₪.
Its fixed costs are 250 ₪ a day.
The profit a day is $\displaystyle {P(x)}={x(100-2x)-(250+20(100-2x))}$, where $\displaystyle x$ is the price of a cake.

Expand the brackets and collect like terms.

Mark on the graph the vertex of the profit’s parabola, and the points where the profit is 0.

Solve the equation: at which prices is the profit 0?

At what price should the cake shop sell a cake, and what is the profit a day?
At which prices does the cake shop make a loss?

A company sells subscriptions to a fitness app.
Its profit a month, in thousands of shekels, is $\displaystyle {P(x)}={-10x^{2}+400x-3000}$, where $\displaystyle x$ is the monthly price of a subscription in shekels.

Which subscription price gives the greatest profit?

What is the greatest profit a month, in thousands of shekels?

Solve the equation: at which prices is the profit 0?

A board member suggests a price of 35 ₪ a month.
What will happen to the company at that price?
Explain with the help of your answers above.

At the cake shop of Exercise 1, the revenue a day is $\displaystyle {R(x)}={x(100-2x)}={-2x^{2}+100x}$.

Which cake price gives the greatest revenue?

Why is the most profitable price, which you found in Exercise 1, higher?