A manufacturer models its daily profit, in dollars, from producing hundred units with the function for . According to this model, what is the maximum possible daily profit, in dollars?

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SAT Nonlinear Functions Question 95 | Free SAT Prep | Aniko

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Question 95·Medium·Nonlinear Functions

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A manufacturer models its daily profit, in dollars, from producing $x$ hundred units with the function

P(x) = -4(x - 6)^2 + 392

for $0 \le x \le 12$ . According to this model, what is the maximum possible daily profit, in dollars?

When a quadratic modeling a real-world quantity is given in vertex form $a(x-h)^2 + k$ , quickly note the sign of $a$ to decide if the vertex is a maximum (if $a<0$ ) or minimum (if $a>0$ ). Then read off $h$ as the input giving the extreme value and $k$ as the extreme value of the function—checking that this $x$ lies within any given domain; if it does, the answer is the vertex’s $y$ -value, otherwise compare function values at the interval endpoints.

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Identify the type and form of the function

Look at the structure of $P(x) = -4(x - 6)^2 + 392$ . Is this a line or a parabola? Is it in a special form you recognize?

Think about maximum vs. minimum

The coefficient of $(x-6)^2$ is negative. For a parabola, what does a negative leading coefficient tell you about whether the vertex is a maximum or a minimum?

Locate the vertex and interpret it

In the form $a(x-h)^2 + k$ , the vertex is at $(h,k)$ . Use this to find which value of $x$ gives the extreme profit, then plug that $x$ back into $P(x)$ to get the profit itself.

Desmos Guide

Graph the profit function with the domain restriction

In Desmos, type y = -4(x - 6)^2 + 392 {0 <= x <= 12} so the graph only shows the relevant part of the parabola.

Find the maximum value on the graph

Tap on the highest point of the parabola (the vertex) or use the function analysis tools to locate the maximum; read off the corresponding y-value, which is the maximum daily profit.

Step-by-step Explanation

Recognize the form and what it tells you

The given function is

P(x) = -4(x - 6)^2 + 392.

This is a quadratic in vertex form, $a(x-h)^2 + k$ . Here, $a = -4$ , $h = 6$ , and $k = 392$ .

Because $a=-4$ is negative, the parabola opens downward, so the vertex represents a maximum value of $P(x)$ .

Find the x-value where the maximum occurs

In vertex form $a(x-h)^2 + k$ , the vertex is at $(h,k)$ . That means this function has its vertex at $x = h = 6$ .

So the maximum profit occurs when $x = 6$ (that is, when the company produces 6 hundred units).

To find the corresponding profit, substitute $x = 6$ into $P(x)$ :

P(6) = -4(6 - 6)^2 + 392.

Compute the maximum profit

Now simplify the expression for $P(6)$ :

First compute the squared term: $(6 - 6)^2 = 0^2 = 0$ .
Then multiply by $-4$ : $-4 \times 0 = 0$ .
Finally add 392:

P(6) = 0 + 392 = 392.

So, according to the model, the maximum possible daily profit is 392 dollars.

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Step-by-step Explanation

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Desmos Guide

Step-by-step Explanation