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Why did Caryl Churchill bother?

Yeah, 4 by 4, but I can't explain it either...other than the fact I know it from a non-calculus trick. :\

I'm pretty sure there's some fancy math you can do to prove it involving maximisation and minimisation calculus stuff. But 4 by 4 sounds right to me as well.

I'm doing an essay on gay marriage; whether or not it should be legalized. I can't think of an appropriate title, can anyone help?

Let the two sides equal y and y+a. Therefore the perimeter is 4y+2a. The perimeter is therefore minimised when a is equal to zero, and the sides are y and y; making it a square

I'll check the text book; but why use calculus if there are other ways you can use it.

Considering you are in a Calculus course, and if this question comes up on a test, teacher's will not give you full marks if you solve a problem without using a Calculus related method.

I doubt that question would ever come up on a Calculus exam, because there's no way that any mathemeticiam would do that using Calculus (unless forced to)

What I'm trying to do is we know that y^2 + ay = 16, so if we get that in terms of a or y, we sub it into 4y + 2a = x, then differentiate in terms of x. The problem is getting either in terms of the other. I tried putting it in the quadratic formula, but can't get anything meaningful out.

Sure, there's plenty of math problems one can solve without the use of calculus, but Calculus courses tend to require the use of calculus. At least, my course did. Other methods, though they may be correct, don't count.

Define a rectangle with sides of length x and y. Let P equal the perimeter function.

P = 2x + 2y
Also, xy = 16 -> y = 16/x

Replacing 2y with 32/x in the perimeter function, we get:

P(x) = 2x + 32/x

Taking the derivative of P(x) we get:

P'(x) = 2 + -32/x^2

To minimize the function, set the derivative equal to zero.

2 + -32/x^2 = 0
x = 4 or x = -4

Clearly, -4 is not a possibility as side lengths must always be positive.
Thus, x = 4 will minimize P(x). P(4) = 16, your answer.

Alas, M. Bison has my answer (I foolishly put P = x + y without thinking, which led to my problem in the first place), but thank you all for helping. :)

Same here. But, we at least get summer reading (just two or three books for English, not that bad). By the way, I am in high school

A problem dealing with Critical Points,

3. For each of the following, find the critical points. Use the first derivative test to determine whether the critical point is a local maximum, local minimum, or neither.

a) y = x^4 - 8x^2

I'm stuck; I don't know what to do!

Take the derivative of the function.

f'(x) = 4x^3 - 16x

Critical points occur when the derivative equals zero.

0 = 4x^3 - 16x

Solving this, you find f'(x) equals zero when x = 2, x = 0, or x = -2.

Next, test the points. To test x = 2, choose a value between infinity and 2 and a value between 0 (the closest critical point) and 2. For example, 3 and 1. Plug these values into f'(x) and note the sign of f'(x).

f(1) = -12, negative
f(3) = 60, positive

This means at x = 2, the function's derivative changes from negative to positive. This means the function itself is decreasing from 0 to 2 and then increasing thereafter. Thus, x = 2 represents a local minimum.

Do the same thing for your other critical points.

Well, I did that but they want exact coordinates like (x,y). I got both the local minimum coordinates but I couldn't get the local maximum coordinate.

Can you help me to get that?

The value of the y coordinate is simply the value of the function, f(x) at x.

Thus the critical points are:

x = 2
y = f(2) = -16
(2, -16)

x = 0
y = f(0) = 0
(0, 0)

x = -2
y = f(-2) = -48
(-2, -48)

The point (0, 0) is a local maximum. This can be verified by the first derivative test (done in my previous post).