A music venue models the revenue , in dollars, from selling tickets to a concert as If the venue can seat no more than 25 people, how many tickets must be sold for the revenue to equal \$1,200?

Solution available with step-by-step explanation

SAT Nonlinear Functions Question 83 | Free SAT Prep | Aniko

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Question 83·Medium·Nonlinear Functions

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A music venue models the revenue $R$ , in dollars, from selling $x$ tickets to a concert as

R(x) = -4x^{2} + 160x.

If the venue can seat no more than 25 people, how many tickets must be sold for the revenue to equal $1,200?

For quadratic word problems about revenue or profit, first translate the words into an equation by setting the given function equal to the specified revenue or profit value. Rearrange the equation so one side is 0, simplify (often by dividing out a common factor), and then solve the quadratic by factoring or using the quadratic formula. Finally, check each solution against any real-world constraints in the problem (such as maximum capacity or nonnegative quantities) before matching it to the answer choices.

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Relate the function to the target revenue

You are given a formula for revenue $R(x)$ and told the revenue should be $1{,}200$ . How can you write an equation that connects $R(x)$ and $1{,}200$ ?

Rewrite as a standard quadratic

Once you have your equation, move all terms to one side so that the equation equals $0$ , and then simplify by dividing out any common factor from all terms.

Solve and check the real-world constraint

Solving the quadratic will give you two values of $x$ . Both work algebraically, but you must also consider that the venue can seat at most 25 people. Which solution fits this condition?

Desmos Guide

Graph the revenue function and target revenue

In Desmos, enter y = -4x^2 + 160x as the first equation, and then enter y = 1200 as a second equation to represent the target revenue.

Find the relevant intersection

Use Desmos to find the intersection points of the parabola and the horizontal line (tap on the intersection dots). Note the $x$ -coordinates of these points; the one that is less than or equal to 25 and matches an answer choice is the number of tickets that gives a revenue of $1{,}200$ .

Step-by-step Explanation

Set up the revenue equation

The revenue function is

R(x) = -4x^2 + 160x.

We want the revenue to be $1{,}200$ , so set $R(x)$ equal to $1{,}200$ :

-4x^2 + 160x = 1200.

Move all terms to one side:

-4x^2 + 160x - 1200 = 0.

Simplify the quadratic equation

All the coefficients are divisible by $-4$ , so divide both sides by $-4$ to make the numbers smaller:

\frac{-4x^2}{-4} + \frac{160x}{-4} + \frac{-1200}{-4} = 0

which simplifies to

x^2 - 40x + 300 = 0.

Now you have a simpler quadratic equation to solve.

Solve the quadratic

Factor the quadratic $x^2 - 40x + 300$ . You need two numbers that multiply to $300$ and add to $40$ . Those numbers are $10$ and $30$ , so

x^2 - 40x + 300 = (x - 10)(x - 30) = 0.

Set each factor equal to $0$ :

x - 10 = 0 \quad \text{or} \quad x - 30 = 0,

so the possible solutions are $x = 10$ and $x = 30$ . These are the ticket amounts that algebraically give a revenue of $1{,}200$ .

Apply the seating limit and choose the valid solution

The venue "can seat no more than 25 people," so the number of tickets sold must satisfy $x \le 25$ .

Of the two values $10$ and $30$ , only $10$ is less than or equal to $25$ , so $x = 30$ is not possible in this real situation.

Therefore, the venue must sell $10$ tickets for the revenue to be $1{,}200$ , which corresponds to choice B.

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Desmos Guide

Step-by-step Explanation

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Step-by-step Explanation