dA/dx DONT UNDERSTAND area and shape perimeter HELP PLEASE HELP!!!?

Question

Bscally i can show that the best rectangle to get the most area from same perimeter is square.also trying to find one that i can use to apply to all shapes (to decgon wil do) to show that the regular ones get the biggest area for the same perimeter and iv got something below that ur spsd to b able to use on other shapes PLZ SHOW HOW.

(Rectangle)
Perimeter=P let a Length=X Then a width is 1/2(P-2X)

Area= X ((1/2 (P-2X)
= X ( P/2-X)
= PX/2-X^2

HERE IS THE IMPORTANT BIT BELOW PLZ HELP!!!!!!

dA/dx = P/2 - 2x = 0 for max or min <<<<< So 2x = P/2 and x = P/4. - I understand how to get from the above but i dont understand the dA/dx thing

please explain, iv luked at some others and they talk about derivates or something and going from 100-2X to 100X-X^2 + C where C is a constant.

But i dont understand how theyve got to there and how on example earlier they got to that either.

ALL HELP GR8ly APPRECIATED

Answer 1

For a rectangle, all that is going on is P is being viewed as a constant number. Because of this, you can imagine P as any number you like (and as you hopefully know, the derivative of a constant is zero). So, when you take the derivative of area (A) you get:

dA/dx = d/dx ( 0.5*P*x - x^2 ) ... 0.5*P is a constant number, so that does not change. Derivative of x is one, so first part is 0.5P. Second part is -x^2 ... short cut- take the two in front, subtract 1 from the exponent and you get -2x. Whole thing together is dA/dx = 0.5*P - 2x. The constant C shouldn't show up because unknown constants only show up from integration, not differentiation.

Doing this for other shapes will require you to rederive each equation. Though you will probably need more information than just one side to do pentagons, hexagons, etc.

Answer 2

dA/dx is basic calculus. If you haven't studied calculus, then that's your fundamental problem.

If F is any function of x, then dF/dx is the derivative of F with respect to x. dF/dx is a function which gives the slope of the graph of F(x) vs. x.

If F is a smooth function, then when it reaches a (local) maximum or minimum, the hill or valley has a flat top or bottom; at the maximum or minimum, the slope, that is dF/dx, is 0.

Here we have a function A giving the area of a rectangle with side x and perimeter P. P is fixed, so we consider A to be a function of x. We want to find the value of x which maximizes A. Since A is a smooth function of x, at that maximum, dA/dx will be 0.

That is why we are looking for an x such that dA/dx = 0.

As you've already noted, with A = (1/2)X(P-2x) = (1/2)(Px - 2x^2) we have:
dA/dx = (1/2)(P - 4x)

for dA/dx to be 0, we have to have P = 4X or x = P/4

As to the computation of dA/dx itself, there are three rules involved:

d(A+B)/dx = dA/dx + dB/dx so
d(Px - 2x^2)/dx = d(Px)/dx + d(-2x^2)/dx

d(kA)/dx = k(dA/dx) so
d(Px)/dx = P(dx/dx) and d(-2x^2)/dx = (-2)d(x^2)/dx

d(x^n) = nx^(n-1) so
d(x^2) = 2x

(remember x^0 = 1, x^1 = x, etc.)

So, if you still don't understand it, which part do you not understand?