Rectangles, Squares, Areas, and Calculus

Rectangles, squares, areas, and calculus are closely related, but do you know how much? Well, see this question. It's very common with numbers, I'm just bringing the general version.

A rectangle with sides of lengths x and y has perimeter P. Prove that the largest area of this rectangle is given when the rectangle is a square.

Well, the question already gives us the variables. All we need is to use them, combine all four terms in the question (rectangles, squares, areas, and calculus) and prove what we need to prove.

So, the perimeter is P. This means that 2(x + y) = P. To get an equation with one variable, let's find y in terms of the other numbers:
x + y = P/2
y = P/2 - x

Now we know two sides of the rectangle: x and P/2 - x. Let's multiply them to find the area function:
A = x(P/2 - x)
A = xP/2 - x²

To find the maximum value for A, let's differentiate:
A' = P/2 - 2x

Equate A' to zero:
0 = P/2 - 2x
2x = P/2
x = P/4

To make sure it's a maximum, let's find the second derivative of A and see if it's negative:
A'' = -2

That's definitely negative, so x = P/4 is the point where the area is maximal. As you can see, that's a quarter of the perimeter. When one side of a rectangle is a quarter of the perimeter, it's a square. To prove that, plug P/4 for x in the equation for y:
y = P/2 - P/4
y = P/4

It's now proven. The biggest area of a rectangle with a constant perimeter is a square.
Nadav

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Inequality Induction

Induction is an untouched topic in this blog so far, but I finally found a good question about it. It's not hard, but it's rather challenging.

Prove by induction:
(1 * 3 * 5 * 7 * ... * (2n - 1)) / (1 * 2 * 3 * 4 * ... * n) < 2ⁿ/√(2n + 1)

The first step in proving an induction is checking whether it works at all. Let's plug 1 for n:
1 / 1 < 2¹/√(2 + 1)
1 / 1 < 2 / √3 - correct!

Assume that the inequality is correct for n = k (k is natural)
(1 * 3 * 5 * ... * (2k - 1)) / (1 * 2 * 3 * ... * k) < 2^k/√(2k + 1)

Now let's prove that if the inequality is right for n = k, it is right for n = k + 1:
(1 * 3 * 5 * ... * (2k - 1) * (2k + 1)) / (1 * 2 * 3 * ... * k * (k + 1)) < 2^{k + 1}/√(2k + 3)

First, we know that (1 * 3 * 5 * ... * (2k - 1)) / (1 * 2 * 3 * ... * k) is smaller than 2^k/√(2k + 1), so it's definitely smaller than 2^{k + 1}/√(2k + 3). For this reason, we can place 2^k/√(2k + 1) instead of this big expression:
(2^k * (2k + 1))/(k + 1)√(2k + 1) < 2^{k + 1}/√(2k + 3)

Divide by 2^k:
(2k + 1)/(k + 1)√(2k + 1) < 2 / √(2k + 3)

Square both sides and cross multiply (k is natural, all positive):
4k² + 8k + 3 < 4k² + 8k + 3
3 < 4 - always true

We have shown that if n = k is correct, n = k + 1 is also correct. By checking that n = 1 is correct, we have proven that the inequality is correct for all natural numbers.

Enjoy,
Nadav

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Parabolic Minimums

Today's question deals with parabolas, lines, and some calculus. It's not very hard, but, as always, it's very interesting.

Two parabolas, y = x²/x + 7 and y = -x²/4 + 3x, are drawn on the same coordinate plane. A vertical line is going through the parabolas. P and Q are the points of intersections with the first and second parabolas, respectively. Find the vertical line and the shortest distance between P and Q.

The first thing to do when you face such a question is to find an equation that represents the quantity you want, this time the distance PQ. Since the two points are on the same vertical line, they both have the same x-coordinate. Let's call it x. That makes the points:
P(x, x²/2 + 7)
Q(x, -x²/4 + 3x)

The distance between them is just the difference in the y-coordinates (because they lie on a vertical line). The distance is:
d = x² + 7 - (-x²/4 + 3x) = x²/2 + 7 + x²/4 - 3x = 3x²/4 - 3x + 7

We want the shortest d possible, and that's achieved when the derivative of d is zero:
d' = 3x/2 - 3
0 = 3x/2 - 3
3 = 3x/2
6 = 3x
x = 2

Now we know the vertical line, x = 2. Plug this value in d and find the shortest distance between P and Q:
d = 3 * 2²/4 - 3 * 2 + 7 = 3 - 6 + 7 = 4

The shortest distance between P and Q is 4.

Have a great week,
Nadav

nadavs

Logarithms and Inequalities

Being independent from Yahoo Answers feels great. I can go to my source, find a great question, and post it here. The topic are somewhat repeating, but there are many more questions. Today's question is about logarithms and inequalities.

Solve:
x^{log₂ x + log_1/4 4} < 4

The solution here is very simple, yet it requires some mathematical knowledge, like what is log_1/4 4. Well, the answer is -1. When the base and the argument are reciprocals, the logarithm's value is -1. Now we can really start solving.

x^{log₂ x - 1} < 4

As you can see, there is a log with base 2 in the exponent. To get rid of the exponent, let's take log base 2 from each side. Since the base is bigger than 1, the inequality sign remains as it is.
log₂ x^{log₂ x - 1} < log₂ 4

Using the log property that says log_a xⁿ = nlog_a x, we can say that:
(log₂ x - 1)log₂ x < 2

Let t = log₂ x

(t - 1)t < 2
t² - t < 2
t² - t - 2 < 0
(t - 2)(t + 1) < 0
-1 < t < 2

Plug the real value of t back:
-1 < log₂ x < 2

By the definition of logs:
1/2 < x < 4

Hope you liked it,
Nadav

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