Maximizing School Profit

Today's question is dealing with schools. As you know, schools need money, so they send dance tickets. However, they need to know how much to charge. Here is a question dealing exactly with that.

A school is holding a dance with a ticket price of $6. 250 plan to attend. The dance team knows that for every $1 increase in the ticket price, 25 less people will come. How much should a ticket cost for maximum revnue? Solve using two different methods.

First, we need an equation to solve, otherwise we won't be able to solve it even with one method. As we know, for each $1 price increase, 25 less people will come. So let x be the number of $1 price increases.

When you increase the price x times, the price goes up by $x ($1 * x). When the price is increased x times, 25x less people will come. The profit is the number of people times the price per ticket. Since the starting price is 6, one factor will be (6 + x). Since 250 plan to attend initially, the second factor is (250 - 25x). This makes the entire profit function:
P = (x + 6)(250 - 25x)
P = -25x² + 100x + 1500

To solve that, we can use two different methods. The first one is using vertex form, which will give us the x coordinate of the vertex, and reveal the maximum point that way (it's an upside-down parabola).
P = -25x² + 100x + 1500
P = -25(x² - 4x - 60)

Complete the square inside:
P = -25(x² - 4x + 4 - 64)
P = -25(x - 2)² + 1600

This means the maximum profit is generated when x = 2 and the ticket price is $8. In this case, 200 people will come, and the profit will be $8 * (250 - 25*2) = $8 * 200 = $1600.

To solve the other way, differentiate P:
P' = -50x + 100

Compare to zero:
0 = -50x + 100
50x = 100
x = 2

Once again, the same result.

Hope you liked it,
Nadav

nadavs

Wire Optimization

Finally, after waiting for a good question, it finally came: an optimization question. This one is from Yahoo Answers, and believe me, it's very good.

A 6 meter long wire is cut into 12 pieces. From these pieces, eight have the same length and the other four also have an equal length. These pieces form a frame of the box. How long should each piece for the box to have a maximum volume?

First, we should define ourselves some variables, otherwise we'd be lost.

Let x be the length of one of the 8 pieces. That means the total length of the 8 equal pieces is 8x. That means the other four pieces have a total length of 6 - 8x, and 3/2 - 2x meters each.

Since these wires form a box, the sides of the box have lengths of x, x, and 3/2 - 2x. That makes the volume of the box x * x * (3/2 - 2x), or 3x²/2 - 2x³.

To find the maximum value, we need to differentiate the volume function and set it to zero. Then we need to find the values that we get from solving the equation, plug them in the second derivative, see which one is a maximum and say "Problem Solved".

So:
V = 3x²/2 - 2x³
V' = 3x - 6x²
0 = 3x - 6x²
3x(1 - 2x) = 0
x = 0, 1/2

Now, let's find the second derivative and see when it's negative, so we get a maximum:
V'' = 3 - 12x
V''(0) = 3 - 0 = 3 > 0 - minimum
V''(1/2) = 3 - 12 * 1/2 = 3 - 6 = -3 < x =" 1/2" 1 =" 1/2">3 = 1/8 m³

Problem solved.
Nadav

nadavs