Farmer brown, just hire a contractor already.

[1MileCrash]

Homework Statement

Suppose that farmer brown has 260 feet of fencing and wants to build a pen that adjoins the whole side of his 100-ft barn (shown). What should the dimensions be for maximum area?

Homework Equations

The Attempt at a Solution

Alright, sorry no picture, but essentially we have a perimeter of 260, one of the sides must be at least 100 ft (adjoins the whole barn) and we want maximum area.

Well, my immediate thought it that a square always has the most area for any rectangle with a given perimeter, so I'm assuming we want what's AS CLOSE to a square as possible, which would be 100X30, yielding an area of 3000 square feet. But I suppose I need to prove this with calculus.

So.. any suggestions?

If perimeter is 260:
260 = 2x + 2y

y = 130 - x

Okay, y = 130 - x, and y must be at least 100. Y is greater than 100 when x is less than 30, so my answer is 30 because that's closest to a square, but let's prove it with calculus..

A = xy = x(130-x)
= 130x - x^2

A' = 130 - 2x

Okay, so now I need to find all critical points of the derivitive of area, where x =< 30, right?

so immediately, I have a critical point of 30, because my interval is (-infinity, 30]

Solve the derivative for 0,

0 = 130 - 2x
130 = 2x
65 = x

So we have a critical point of 65 (perfect square) and 30, but 65 is not in our interval, therefore the only critical point is 30, therefore 30 is the maximum within the interval, therefore the dimensions for maximum area are 100 x 30.

Is this adequate proof?

 

[micromass]

Hi 1MileCrash!

That's a good approach, but there is one (very minor gap) here. You said that 30 was a critical point, and that's certainly correct. But this only implies that 30 can be a maximum or a minimum. You must show that it cannot be a minimum. (because if it were a minimum, then any other value is better than 30!)

 

[1MileCrash]

Hi 1MileCrash!

That's a good approach, but there is one (very minor gap) here. You said that 30 was a critical point, and that's certainly correct. But this only implies that 30 can be a maximum or a minimum. You must show that it cannot be a minimum. (because if it were a minimum, then any other value is better than 30!)

Solve A' for any point on the interval (-infinity, 30] and confirm it is positive?

A' = 130 - 2(29)
A' = 72

Therefore, since there are no other critical points on the interval, it has a positive slope the entire way, so it is always increasing on the interval, so of course, 30 is the maximum?

 

[micromass]

Solve A' for any point on the interval (-infinity, 30] and confirm it is positive?

A' = 130 - 2(29)
A' = 72

Therefore, since there are no other critical points on the interval, it has a positive slope the entire way, so it is always increasing on the interval, so of course, 30 is the maximum?

Yes, that sounds good. I may seem stupid now to check this, but sometimes it is important not to forget it.

 

[1MileCrash]

Yes, that sounds good. I may seem stupid now to check this, but sometimes it is important not to forget it.

No, I completely understand its importance for proof's sake.

Thanks!

 

[SammyS]

I would assume that the side of the pen adjoining the barn doesn't need any fence !

If we assume a rectangular shape AND you MUST use the whole side of the barn, then you have 260 = 2x + y + (y -100), and y ≥ 100.