Can someone explain Optimization in Calculus?

Question

Before you answer this question, please note that if I understand the concept of optimization, then I won't need you to actually do the problem I'm posting. If it's easier for you to explain while answering, that's alright too, but I really need to understand the steps of optimization.

What I don't understand is how to determine the maximum or minimum area/perimeter/volume of a figure and how to determine the area/perimeter/volume of certain figures as a one variable function and then solve that function from there. I can only think of one good example, so I'll update later if I think of another one.

1) Consider a fence 80 meters long. Suppose you are building a rectangular garden with a missing side. Determine the maximum size garden you can construct.

The first thing I did was sketch a diagram. Then, I wrote down all that I knew, namely that the perimeter is 80m, and the formula for this is 2l + w = P (or 2w + l = P).

Then I was asked to determine the area as a one variable function. Since:
area = A = lw
and:
P =80 = 2l + w
I said that:
l = 40 - (w/2)
Therefore:
A = 40w - ([w^2]/2)

If all of this is correct, how do you find the maximum area? How would you find the minimum area?

Answer 1

That's exactly correct. If you plotted A vs w you'd see a parabola. Because the coefficient on w^2 is negative, this is a parabola that opens down. It has a maximum, but no minimum. The parabola goes to -infinity. But since area can't be negative, I guess you'd consider the minimum area to be 0, which is going to happen when the fence has no width or no length.



Normally you don't need fine points such as that on problems like this. All you're supposed to do is take the derivative and find where it is 0. That could either be a maximum or a minimum and you can't tell from the first derivative, but the problem is going to be set up so the place where the derivative is 0 is the right place.



In the real world of optimization, you do need to know which it is, which you get by checking other things including the second derivative.



So in this case, dA/dw = 40 - 2w/2 = 40 - w. Setting this equal to 0, 40 - w = 0 gives w = 40.

That is the optimum value. Plug that into the perimeter formula to find the length.

Answer 2

All of what you've done is perfectly correct. The goal of optimization is to take the derivative of a function so you can find out when the slope is zero. This place marks where it is at a maximum or a minimum. So take the derivative of the equation you came up with. A'=40-w. Set A' equal to zero because you're trying to find when the slope is zero because that's where the (local) maximum is at. 0=40-w. So w=40. Which means that l=20. This is the maximum, I'm not sure how to find the minimum but I hope this helped.

Answer 3

So far you're doing good. Now just take the first derivative of the area equation you came up with:

A ' = 40 - w



Set this equal to zero and solve for w:

0 = 40 - w

w = 40



Now that you know what the width is, you can go back and solve for the length:

80 = 2l + w

80 = 2l + 40

40 = 2l

20 = l



So really the length is 40 and the width is 20 just because length is longer:

40m X 20m <==Answer

Answer 4

a million) it is basic; ==> First come to a decision the point of the priority; right here it is paper dimensions, so as that its area would be minimum; then next comes decrease than what constraints this must be accomplished? ==> nicely, right here on the chosen paper, you're leaving some margins throughout and arriving on the area for printing, this is given some mounted fee. 2) Now, we are sparkling what we prefer; a thank you to proceed for development mathematical equations, so as that it is solved for identifying to purchase the tip answer: 3) right here we are given the printing area as 50 squarewherein is continuing; subsequently enable us to start from this documents; ==> permit the measurement of the printing area be x in (height smart) via y in (width smart) ==> Printing area = xy = 50; == y = 50/x -------(a million) 4) there's a margin of four in each and each at precise and backside are provided; subsequently ordinary height of the paper is "x + 8" in; in addition a margin of two in is provided on the two facets; ==> ordinary width = y + 4 in 5) subsequently the area of the paper is = (x+8)(y+4) 6) Substituting for y from equation (a million), A (x+8)(50/x + 4) = 50 + 4 hundred/x + 4x + 32 7) subsequently the function to be minimized is: A = 80 two + 4x + 4 hundred/x 8) Differentiating this, A' = 0 + 4 - 200/x^2 = 4 - 4 hundred/x^2 9) Equating A' = 0, x^2 = a hundred; ==> x = +/- 10 in 10) yet a measurement can not be destructive, subsequently we evaluate purely + 10; So x = 10 in and y = 5 in 11) despite the fact that we could desire to make certain,whether it is minimum or optimum; for which we can observe 2d spinoff attempt; So lower back differentiating, A'' = 800/x^3; at x = 10, A'' is > 0; subsequently it is minimum subsequently we end for the printing the area to be 50 squarein, decrease than the given constraints of margins, the outer length of the paper must be 18 in via 9 in; So outer area is = 162 squarein. choose you're defined; have a advantageous time.