Math  /  Algebra

Question3) ( 6 marks) The SAC at North Park has decided to sell water bottles to raise money. The demand function is described by p(x)=5x+33p(x)=-5 x+33, where xx is the number of water bottles, in hundreds, and p(x)p(x) is in hundreds of dollars. The cost function, in hundreds of dollars, is described by C(x)=3x+4C(x)=3 x+4. Determine the number of water bottles they must sell so that they maximize their profit.

Studdy Solution

STEP 1

What is this asking? How many water bottles does the North Park SAC need to sell to make the most money? Watch out! Don't forget the demand and cost are in hundreds of dollars, and xx represents hundreds of water bottles!

STEP 2

1. Define the profit function
2. Find the derivative of the profit function
3. Find the critical points
4. Determine the maximum

STEP 3

Alright, let's **start** by remembering how profit, revenue, and cost are related.
Profit is the money left over *after* you've paid your costs, right?
So, Profit = Revenue - Cost.

STEP 4

We know the cost function C(x)=3x+4C(x) = 3x + 4, but what about revenue?
Revenue is how much money you bring in *before* considering costs.
It's simply the price per water bottle times the number of water bottles sold.
In our case, that's p(x)xp(x) \cdot x.

STEP 5

Let's **calculate** the revenue function R(x)R(x): R(x)=p(x)x=(5x+33)x=5x2+33x.R(x) = p(x) \cdot x = (-5x + 33) \cdot x = -5x^2 + 33x.

STEP 6

Now, we can **define** our profit function, P(x)P(x): P(x)=R(x)C(x)=(5x2+33x)(3x+4)=5x2+30x4.P(x) = R(x) - C(x) = (-5x^2 + 33x) - (3x + 4) = -5x^2 + 30x - 4. So, P(x)=5x2+30x4P(x) = -5x^2 + 30x - 4 is our **profit function**!

STEP 7

To maximize profit, we need to find where the profit function peaks.
This happens when the *slope* of the profit function is zero, which means we need to find the **derivative**!

STEP 8

Using the power rule, we **find** the derivative of P(x)P(x): P(x)=10x+30.P'(x) = -10x + 30.

STEP 9

**Critical points** occur where the derivative is zero or undefined.
Our derivative is a nice, simple linear function, so it's defined everywhere.
Let's set it equal to zero and solve for xx: P(x)=10x+30=0.P'(x) = -10x + 30 = 0.

STEP 10

Subtracting 30 from both sides gives 10x=30-10x = -30.

STEP 11

Dividing both sides by -10 gives x=3010=3x = \frac{-30}{-10} = 3.
So, x=3x = \textbf{3} is our **critical point**!

STEP 12

We found a critical point, but is it a maximum?
Let's think about the shape of our profit function, P(x)=5x2+30x4P(x) = -5x^2 + 30x - 4.
It's a parabola that opens downwards (because the coefficient of x2x^2 is negative).
This means our critical point *must* be a maximum!

STEP 13

So, the SAC needs to sell x=3x = \textbf{3} hundred water bottles to maximize their profit.
That's 300\textbf{300} water bottles!

STEP 14

The North Park SAC needs to sell 300 water bottles to maximize their profit.

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