Sinusoid Temperature

Temperatures are very hard to predict. Even weathermen get them wrong most of the times. However, there are some paradises which have a constant temperature range, such as Brazzaville (which is in Congo). Here is the question with the amazing details:

The temperature in Brazzaville is given by the following function:
C = 17 - 6cos(πt/12 + 5)
0 <= t < 24

1. What are the highest and lowest temperatures on Brazzaville?

2. What is the biggest rate of change (biggest drop/rise) in temperature?

To show how people complicate questions, I'm going to answer it using calculus, and then using a much simpler method.

The highest and lowest temperatures are achieved when the derivative of the function is 0, so let's do it:
C' = 6π/12 * sin(πt/12 + 5)
0 = πsin(πt/12 + 5) / 2

Divide by π / 2:
0 = sin(πt/12 + 5)
πt/12 + 5 = π*k (where k is an integer)
πt + 60 = 12πk
πt = 12πk - 60
t = 12k - 60/π

Now all we need to do is find the k values which give t's that match our requirement. Since 60/π is about 19, we need k's that are bigger than 1, so:
t₁ = 24 - 60/π = 4.901
t₂ = 36 - 60/π = 16.901

To verify which number is a maximum and which is a minimum, we can use the second derivative:
C'' = 6π²/144 * cos(πt/12 + 5)

Plug the t's:
t₁ makes it positive, so it's a minimum.
t₂ makes it negative, so it's a maximum.

Plug the two t's in the original equation:

C = 17 - 6cos(πt₁/12 + 5) = 17 - 6 = 11°C
C = 17 - 6cos(πt₂/12 + 5) = 17 - 6(-1) = 23°C

The temperature in Brazzaville ranges between 11°C - 23°C for the entire year. We can only be jealous of them...

To get the biggest rates of change, we need to find the maximum of the derivative, so we'll set the second derivative to zero and find the two points. One will give out a negative change and the other one will give a positive change. They are opposites of each other, since the cosine function is symmetrical on a vertical line going through a maximum or minimum.
C'' = 6π²/144 * cos(πt/12 + 5)
0 = 6π²/144 * cos(πt/12 + 5)
cos(πt/12 + 5) = 0
πt/12 + 5 = ±π/2 + 2πk (again, k is an integer)
πt + 60 = ±6π + 24πk
πt = ±6π + 24πk - 60
t = ±6 + 24k - 60/π

The t values in bounds are:
t₃ = 10.901
t₄ = 22.901

Plug them in C':
C' = π/2 * sin(πt₃/12 + 5) = π/2 = 1.57 °C/hour
C' = π/2 * sin(πt₄/12 + 5) = -π/2 = -1.57 °C/hour

So the sharpest temperature changes are 1.57 °C per hour.

This question can be solved using a much easier way: common sense.

The t inside the function is multiplied by π and divided by 12, so plugging in t values from 0 to 24 will give values from 0 to 2π inside the cosine function. This means it goes through an entire cycle from t = 0 to t = 24. With that, we can figure out that the maximum value for the cosine function is 1 and the minimum value is -1. Plug these two numbers and you get:
C = 17 - 1*6 = 11°C
C = 17 - (-1)*6 = 23°C


Also, the cosine function is the "steepest" when it goes through the x-axis, or more accurately, when the argument makes the cosine zero. All we need to do is find when the argument is zero and plug it in the derivative:
πt/12 + 5 = ±π/2 + 2πk

I already solved that, so I'm not going to do it again.

As you can see, thinking outside the box can really help when it comes to such long and tiring functions.

Yours,
Nadav

nadavs

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

Boating Calculus

Today's question is dedicated to all boating lovers and calculus fans. It is not very hard, it has no use in real life, and yet it is a very beautiful question (like the ones on super math tips).

A boat leaves the dock at 12:00pm and heads west at 40km/h. Another boat travels north at 20km/h and enters the dock at 1:00pm. When is the distance between the boats the shortest and what is that distance?

Nice, isn't it? We have an optimization question here, so we need to create a function to find a variable to optimize. We're dealing here with given speeds, variable times and varying distances, so let's call the time variable t (in hours, to match the speeds) and the distance variable d (in km).

Now, the distance from the first boat to the dock is 40t (it starts at the dock). The distance from the second boat to the dock is 20 when it starts, and 20 - 20t total (since it travels towards the dock).

The boats are heading in perpendicular directions, so they form a right triangle. Using the Pythagorean theorem, we can find the distance between the boats, d:
d² = (40t)² + (20 - 20t)²
d² = 1600t² + 400 - 800t + 400t²
d² = 2000t² - 800t + 400

Since we want to find the minimum value for d and d must be positive, we can also find the minimum value of d² (also must be positive). Let y = d²:
y = 2000t² - 800t + 400

Now differentiate this function:
y' = 4000t - 800

To find an extremety, we need to set y' to zero:
0 = 4000t - 800
4000t = 800
t = 1/5

To see that it is indeed a minimum, let's find the second derivative of y:
y'' = 4000
Since it is positive, y is minimal when t = 1/5

Now calculate y when t = 1/5:
y = 2000 * (1/5)² - 800 * (1/5) + 400
y = 2000 * (1/25) - 160 + 400
y = 80 + 240 = 320

However, y is the distance squared, so let's take the square root of this number:
d = sqrt(y) = sqrt(320) = 8sqrt(5) = 17.888... km.

This means that after 1/5 of an hour, which is 60 / 5 = 12 minutes, the boats will be the closest to each other at 8sqrt(5) (or 17.888) km apart.

Hope you enjoyed.
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