SAT Nonlinear Functions Question 107 | Free SAT Prep

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Question 107·Medium·Nonlinear Functions

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The revenue, in dollars, from selling a certain product is modeled by

R(p) = -2(p-40)^2 + 3200,

where $p$ is the selling price, in dollars, of the product. According to the model, what price maximizes the revenue?

When a revenue or profit question gives you a quadratic in vertex form, immediately match it to $a(x-h)^2 + k$ . Check the sign of $a$ : if it’s negative, the vertex gives the maximum; if positive, it gives the minimum. The $x$ -coordinate of the vertex is simply $h$ , so you can read it directly from the expression inside the square, being careful about signs. Finally, answer in terms of the variable the question asks about (often price or quantity), not the function value (revenue or profit).

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Identify the function shape

Look at the coefficient in front of $(p-40)^2$ . Is it positive or negative, and what does that tell you about whether the parabola opens up or down?

Use vertex form

The function is written as $R(p) = a(p-h)^2 + k$ . In that form, what does $h$ represent on the graph of the function?

Make the squared part zero

For this function, the squared part is $(p-40)^2$ . What value of $p$ makes $p-40$ equal to $0$ ? That $p$ -value is where the maximum occurs.

Desmos Guide

Graph the revenue function

In Desmos, enter the function as y = -2(x-40)^2 + 3200. This graphs revenue (on the $y$ -axis) as a function of price $x$ .

Find the maximum point

Tap on the highest point of the parabola or use the Desmos maximum feature on the graph. Look at the $x$ -coordinate of this maximum point; that $x$ -value is the price that maximizes the revenue.

Step-by-step Explanation

Recognize the function type and form

The revenue is given by

R(p) = -2(p-40)^2 + 3200.

This is a quadratic function of $p$ and it is already in vertex form:

R(p) = a(p-h)^2 + k,

where the vertex is at $(h, k)$ .

By comparing, you can identify $a = -2$ , $h = 40$ , and $k = 3200$ (but we will use $h$ and $a$ more than $k$ ).

Decide whether the vertex is a maximum or a minimum

The coefficient of the squared term is $a = -2$ , which is negative. That means the parabola opens downward (like an upside-down U).

For a downward-opening parabola, the vertex is the highest point on the graph. In this context, that means the vertex gives the maximum revenue.

So the price that maximizes revenue is the $p$ -value (horizontal coordinate) of the vertex.

Find the price at the vertex

In vertex form $R(p) = a(p-h)^2 + k$ , the vertex occurs at $p = h$ .

Here the squared part is $(p-40)^2$ , so it has the form $(p-h)^2$ with $h = 40$ .

Equivalently, the vertex occurs when the squared term is zero:

(p-40)^2 = 0 \implies p - 40 = 0 \implies p = 40.

Therefore, according to the model, the revenue is maximized when the product is sold at a price of $40.

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Desmos Guide

Step-by-step Explanation

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Strategy

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Desmos Guide

Step-by-step Explanation