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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Sphere Differentiation

.diToday's question deals with differentiation and spheres. The question uses some very nice properties of differentiation and the dy/dx notation. It has two parts, but so do most questions from text books (apparently, people at Yahoo Answers aren't that original in making up questions).

A sphere has radius of length r cm, surface area of S cm², and a volume of V cm³ on a given instant, t.

1) Prove: (dV/dt) * (dr/dt) = (dS/dt)² / 16π

2) Find the surface area when dV/dt = π cm³/sec and dS/dt = 2π cm²/sec

Not that bad, is it?

First, there are two important formulas we are going to use in order to solve this problem:
V = 4πr³ / 3
S = 4πr²

Now differentiate each function according to t:
dV/dt = 4πr² dr/dt
dS/dt = 8πr dr/dt

To get dV/dt * dr/dt, multiply the first equation by dr/dt:
dV/dt * dr/dt = 4πr² (dr/dt)²

Now square the second equation:
(dS/dt)² = 64π²r² (dr/dt)²

Divide the second equation by 16π:
64π²r² / 16π = 4πr² (dr/dt)²

Now this is exactly the same as dV/dt * dr/dt.
Q.E.D

The second question:
dS/dt = dS/dV * dV/dt
dS/dV = dV/dt * dS/dt
dS/dV = (8πr dr/dt) / (4πr² dr/dt)
dS/dV = 2/r
dS/dt = 2/r * dV/dt
dS/dt = 2/r * π
dS/dt = 2π
2π = 2π/r
1 = 1/r
r = 1

Plug r = 1 into the surface area function:
S = 4π * 1² = 4π cm²

Problem solved.

Hope you liked it.

Nadav

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