Math / Algebra

Questionmetres above the river. Determine the equation that models the arches. Research shows that for every $\$ 0.10$ increase in the price of batteries, there is a 10 decrease per day. If a store normally sells 600 packs of batteries at $\$ 4.95$ , what is optimum price to sell batteries at?

Studdy Solution

STEP 1

What is this asking? If we raise the price of batteries, we sell fewer batteries.
What's the best price to make the most money? Watch out! Don't get tricked!
We want to maximize the total money made, not how many batteries are sold!

STEP 2

1. Define the variables and the revenue function.
2. Find the vertex of the parabola.
3. Calculate the optimal price.

STEP 3

Let $x$ be the number of $\$0.10$ price increases.
So, if $x = 1$ , the price goes up by $\$0.10$ .
If $x = 2$ , the price goes up by $\$0.20$ , and so on.
Let $R(x)$ be the revenue, which is the total money made from selling batteries.

STEP 4

The initial price is $\$4.95$ .
For every increase $x$ , the price goes up by $\$0.10 \cdot x$ .
So, the new price $P(x)$ is given by: $P(x) = 4.95 + 0.10x$

STEP 5

The initial quantity sold is 600.
For every increase $x$ , the quantity sold goes down by 10.
So, the new quantity sold $Q(x)$ is given by: $Q(x) = 600 - 10x$

STEP 6

Revenue is price multiplied by quantity.
So, our revenue function $R(x)$ is: $R(x) = P(x) \cdot Q(x) = (4.95 + 0.10x)(600 - 10x)$

STEP 7

Let's expand $R(x)$ to get a nice, standard parabola equation: \begin{align} R(x) &= (4.95 + 0.10x)(600 - 10x) \\ &= 2970 - 49.5x + 60x - x^2 \\ &= -x^2 + 10.5x + 2970\end{align} This is a downward-facing parabola because the coefficient of $x^2$ is negative!

STEP 8

The x-coordinate of the vertex, where the revenue is maximized, is given by $x = -\frac{b}{2a}$ .
In our case, $a = -1$ and $b = 10.5$ , so: $x = -\frac{10.5}{2(-1)} = \frac{10.5}{2} = 5.25$

STEP 9

The x-coordinate of the vertex tells us the optimal number of price increases.
Since $x = 5.25$ , the price increase is: $5.25 \cdot \$0.10 = \$0.525$ Rounding to the nearest cent, we get $\$0.53$ .

STEP 10

Now, add this increase to the original price: $\text{Optimal Price} = \$4.95 + \$0.53 = \$5.48$

STEP 11

The optimum price to sell the batteries is $\$5.48$ .

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Questionmetres above the river. Determine the equation that models the arches. Research shows that for every $\$ 0.10$ increase in the price of batteries, there is a 10 decrease per day. If a store normally sells 600 packs of batteries at $\$ 4.95$ , what is optimum price to sell batteries at?

Studdy Solution

STEP 1

What is this asking? If we raise the price of batteries, we sell fewer batteries.
What's the best price to make the most money? Watch out! Don't get tricked!
We want to maximize the total money made, not how many batteries are sold!

STEP 2

1. Define the variables and the revenue function.
2. Find the vertex of the parabola.
3. Calculate the optimal price.

STEP 3

Let $x$ be the number of $\$0.10$ price increases.
So, if $x = 1$ , the price goes up by $\$0.10$ .
If $x = 2$ , the price goes up by $\$0.20$ , and so on.
Let $R(x)$ be the revenue, which is the total money made from selling batteries.

STEP 4

The initial price is $\$4.95$ .
For every increase $x$ , the price goes up by $\$0.10 \cdot x$ .
So, the new price $P(x)$ is given by: $P(x) = 4.95 + 0.10x$

STEP 5

The initial quantity sold is 600.
For every increase $x$ , the quantity sold goes down by 10.
So, the new quantity sold $Q(x)$ is given by: $Q(x) = 600 - 10x$

STEP 6

Revenue is price multiplied by quantity.
So, our revenue function $R(x)$ is: $R(x) = P(x) \cdot Q(x) = (4.95 + 0.10x)(600 - 10x)$

STEP 7

Let's expand $R(x)$ to get a nice, standard parabola equation: \begin{align} R(x) &= (4.95 + 0.10x)(600 - 10x) \\ &= 2970 - 49.5x + 60x - x^2 \\ &= -x^2 + 10.5x + 2970\end{align} This is a downward-facing parabola because the coefficient of $x^2$ is negative!

STEP 8

The x-coordinate of the vertex, where the revenue is maximized, is given by $x = -\frac{b}{2a}$ .
In our case, $a = -1$ and $b = 10.5$ , so: $x = -\frac{10.5}{2(-1)} = \frac{10.5}{2} = 5.25$

STEP 9

The x-coordinate of the vertex tells us the optimal number of price increases.
Since $x = 5.25$ , the price increase is: $5.25 \cdot \$0.10 = \$0.525$ Rounding to the nearest cent, we get $\$0.53$ .

STEP 10

Now, add this increase to the original price: $\text{Optimal Price} = \$4.95 + \$0.53 = \$5.48$

STEP 11

The optimum price to sell the batteries is $\$5.48$ .

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Studdy Solution

STEP 1

What is this asking? If we raise the price of batteries, we sell fewer batteries.What's the best price to make the most money? Watch out! Don't get tricked!We want to maximize the *total money* made, not how many batteries are sold!

STEP 2

1. Define the variables and the revenue function.2. Find the vertex of the parabola.3. Calculate the optimal price.

STEP 3

Let xxx be the number of $0.10\$0.10$0.10 price increases.So, if x=1x = 1x=1, the price goes up by $0.10\$0.10$0.10.If x=2x = 2x=2, the price goes up by $0.20\$0.20$0.20, and so on.Let R(x)R(x)R(x) be the revenue, which is the total money made from selling batteries.

STEP 4

The initial price is $4.95\$4.95$4.95.For every increase xxx, the price goes up by $0.10⋅x\$0.10 \cdot x$0.10⋅x.So, the new price P(x)P(x)P(x) is given by: P(x)=4.95+0.10xP(x) = 4.95 + 0.10xP(x)=4.95+0.10x

STEP 5

The initial quantity sold is **600**.For every increase xxx, the quantity sold goes *down* by **10**.So, the new quantity sold Q(x)Q(x)Q(x) is given by: Q(x)=600−10xQ(x) = 600 - 10xQ(x)=600−10x

STEP 6

Revenue is price multiplied by quantity.So, our revenue function R(x)R(x)R(x) is: R(x)=P(x)⋅Q(x)=(4.95+0.10x)(600−10x)R(x) = P(x) \cdot Q(x) = (4.95 + 0.10x)(600 - 10x)R(x)=P(x)⋅Q(x)=(4.95+0.10x)(600−10x)

STEP 7

Let's expand R(x)R(x)R(x) to get a nice, standard parabola equation: \begin{align*} R(x) &= (4.95 + 0.10x)(600 - 10x) \\ &= 2970 - 49.5x + 60x - x^2 \\ &= -x^2 + 10.5x + 2970\end{align*} This is a downward-facing parabola because the coefficient of x2x^2x2 is negative!

STEP 8

The x-coordinate of the vertex, where the revenue is maximized, is given by x=−b2ax = -\frac{b}{2a}x=−2ab​.In our case, a=−1a = -1a=−1 and b=10.5b = 10.5b=10.5, so: x=−10.52(−1)=10.52=5.25x = -\frac{10.5}{2(-1)} = \frac{10.5}{2} = 5.25x=−2(−1)10.5​=210.5​=5.25

STEP 9

The x-coordinate of the vertex tells us the optimal number of price increases.Since x=5.25x = 5.25x=5.25, the price increase is: 5.25⋅$0.10=$0.5255.25 \cdot \$0.10 = \$0.5255.25⋅$0.10=$0.525 Rounding to the nearest cent, we get $0.53\$0.53$0.53.

STEP 10

Now, add this increase to the original price: Optimal Price=$4.95+$0.53=$5.48\text{Optimal Price} = \$4.95 + \$0.53 = \$5.48Optimal Price=$4.95+$0.53=$5.48

STEP 11

The optimum price to sell the batteries is $5.48\$5.48$5.48.

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Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

What is this asking? If we raise the price of batteries, we sell fewer batteries.
What's the best price to make the most money? Watch out! Don't get tricked!
We want to maximize the total money made, not how many batteries are sold!

1. Define the variables and the revenue function.
2. Find the vertex of the parabola.
3. Calculate the optimal price.

Let $x$ be the number of $\$0.10$ price increases.
So, if $x = 1$ , the price goes up by $\$0.10$ .
If $x = 2$ , the price goes up by $\$0.20$ , and so on.
Let $R(x)$ be the revenue, which is the total money made from selling batteries.

The initial price is $\$4.95$ .
For every increase $x$ , the price goes up by $\$0.10 \cdot x$ .
So, the new price $P(x)$ is given by: $P(x) = 4.95 + 0.10x$

The initial quantity sold is 600.
For every increase $x$ , the quantity sold goes down by 10.
So, the new quantity sold $Q(x)$ is given by: $Q(x) = 600 - 10x$

Revenue is price multiplied by quantity.
So, our revenue function $R(x)$ is: $R(x) = P(x) \cdot Q(x) = (4.95 + 0.10x)(600 - 10x)$

Let's expand $R(x)$ to get a nice, standard parabola equation: \begin{align} R(x) &= (4.95 + 0.10x)(600 - 10x) \\ &= 2970 - 49.5x + 60x - x^2 \\ &= -x^2 + 10.5x + 2970\end{align} This is a downward-facing parabola because the coefficient of $x^2$ is negative!

The x-coordinate of the vertex, where the revenue is maximized, is given by $x = -\frac{b}{2a}$ .
In our case, $a = -1$ and $b = 10.5$ , so: $x = -\frac{10.5}{2(-1)} = \frac{10.5}{2} = 5.25$

The x-coordinate of the vertex tells us the optimal number of price increases.
Since $x = 5.25$ , the price increase is: $5.25 \cdot \$0.10 = \$0.525$ Rounding to the nearest cent, we get $\$0.53$ .

Now, add this increase to the original price: $\text{Optimal Price} = \$4.95 + \$0.53 = \$5.48$

The optimum price to sell the batteries is $\$5.48$ .