Question

Help?

find the length and width of a rectangle with area of 50 meters and a maximum perimeter

Answer 1

Well I know P= 2L + 2W
that turns out to: 2(x+y)

Answer 2

Hola, Vibarguen! Buenos Dias! What's the formula for the AREA of this rectangle?

Answer 3

Think for a moment. Which of these two expressions do you want to MAXIMIZE?

As soon as I'd typed MAXIMIZE, I realized that it's much more common to MINIMIZE the perimeter. Sure that you've copied the problem correctly?

Answer 4

Morning ;) A=XY
yeah mathmale, I know! I've done the minimize before but I don't know how to maximize

Answer 5

Take a look:
A = 64 = xy
P = 2(x + y)

P = 2x + 128/x
dP/dx = 2 - 128/x^2
d^2P/dx^2 = 256/x^3
x^2 = 64
dP/dx = 0
x = 8
Relative min

y = 64/x = 8

8 ft x 8 ft

Answer 6

Are you familiar with the Second Derivative Test?

Answer 7

yes

Answer 8

oh and the problem for that was: Find the length and the width of a rectangle that has an area of 64 ft^2 and a minimum perimeter

Answer 9

You're certainly on the right track, and Iappreciate your typing out your own work.
I immediately noticed that you've typed XY=64. Mind checking that against the original problem?

Answer 10

Oh. A different problem. OK.
For all practical purposes, the current problem is very similar to the previous one.

Answer 11

I notice you found the 2nd derivative and ended up with +256/(x^3). For positive x, that means your second derivative is always positive itself. Positive second derivative tells you what about the critical point? Is that point a max or a min?

Answer 12

:)

Answer 13

It tells me it is a min

Answer 14

Right. My point, exactly. So, if x = Sqrt(50) (in the current problem), your perimeter will be at its minimum (not maximum).

So, go back to the formula A=x*y.
We know that both x and y must be greater than zero. If you're in doubt, solve A=xy for y and graph the resulting equation. The graph may help you to answer the question posed here.

Answer 15

so I do the derivative of sqrt 50 ?

Answer 16

because with this one I am honestly not sure how to do it

Answer 17

Not quite. Instead, please go back and review what you did in the previous problem. You were right on target there. You found that at x=8, your perimeter will be at its nimimum.

My point here is the in the current problem, you are doing exactly the same thing. You have the constraint equation xy=50, or y=50/x.

Substitute that into P=2(x+y). Then vary x and determine whether or not the P has a maximum (as opposed to a minimum).

Answer 18

Hint: P = 2x + 100/x. Try graphing that.