Please help me with this one also...

The perimeter of a window, made in the shape of a semicircle on a rectangle, is 4m. Find its dimensions if the area is to be a maximum?

The perimeter is given.

The circumference of a circle is given by . But you have a semi-circle. So,

Let the perimeter of the rectangular region be 2r+2y.



The area of the window is

A is what you must minimize. Solve the perimeter equation for, say, r in terms of y and sub into the area equation. It will then be in terms of one variable. Differentiate, set to 0 and solve for y.

What would r in terms of y look like, I'm having trouble with that as I get a weird answer. I get y=(4-PIr)/2 - another set of r.

Thanks

You will better of using y in terms of r!

Get the valye of y from the equation for P

Substitute the same in Area.

Then find first derivative of A

Make it zero (for Max/Min) and get values of r.

Select value of r which makes second derivative negative.

Sorry but I still can't get the right answer...

pi r + 2 (r + y) = 4

y = ( 4 - pi r)/2 - r

Substitute this in A = pi r^2/2 + 2ry

Then find first and second derivative of A with respect to r

Show me your steps

:eek: uh I feel dumb??

Yeh terry that's what I got but when I sub it in, the problems begins.

A = pi r^2/2 + 2ry

A = pi r^2/2 + 2r [( 4 - pi r)/2 - r]

= pi r^2/2 + r [ (4 - pi r) - 2r ]

= pi r^2/2 - (pi +2) r^2 + 4r

dA/dr = pi r - (pi + 2) 2r + 4

dA^2/dr^2 = pi - (pi +2) 2 = -pi - 4 (NEGATIVE regardless)

dA/dr = 0

(pi +2) 2r = pi r + 4

pi r + 4r = 4

r = 4/(pi + 4)

This gives value of r.

You can find y now:

y = ( 4 - pi r)/2 - r