Calc Question: need help please.

[PHiuR]

i tried to draw it out real quick...heres the pic...the black boarder around the outside is suppose to be the track around the field totaling 400M


http://pics.bbzzdd.com/users/PHiuR/question2.jpg

the track and field is made up of this image...then the perimeter = 400M. what is the max area of the rectangle area?

any help appreciated... thanks!

2nd question =


tan(x)/1+cos^3(x) interval [2,3]

what is the maximum value.? useing [2,3]

and the values of 'c' with the mean value theorem.? useing [2,3]


^ do you use rolles theorem to find the maximum value?

 

[Shooters]

You know that the area of a rectangle is width x height.

From your picture, the height is obviously going to be the diameter of the semicircles, and the width is going to be 0.5*(400 minus the circumference of the two semicircles).

Using that info, you should be able to get an expression for the area of the rectangle in terms of the radius of the semicircles.

Take the derivative and set it equal to zero to find the stationary point(s).

Then take the second derivative and use the second derivative test to determine if the stationary point is a min, max, or saddle (it should be a negative value which indicates a max).

Finally, plug the answer you got from the first derivative into your area equation, and that will be your max area.

The answer should be A = (20,000/pi) square meters unless I've done something wrong.

 

[PHiuR]

Originally posted by: Shooters
You know that the area of a rectangle is width x height.

From your picture, the height is obviously going to be the diameter of the semicircles, and the width is going to be 0.5*(400 minus the circumference of the two semicircles).

Using that info, you should be able to get an expression for the area of the rectangle in terms of the radius of the semicircles.

Take the derivative and set it equal to zero to find the stationary point(s).

Then take the second derivative and use the second derivative test to determine if the stationary point is a min, max, or inflection (it should be a negative value which indicates a max).

Finally, plug the answer you got from the first derivate into your area equation, and that will be your max area.

The answer should be A = (20,000/pi) square meters unless I've done something wrong.

hey, thanks for replying.

why would you use the width as 0.5?

 

[PHiuR]

im bouts to shi* in my pants. can anyone help at all?

 

[rgwalt]

I get the same answer as Shooters for the area of the rectangle. Area of the rectangle is A=x*d where d is the diameter of the circles. Perimeter of the track is 400 = pi*d+2x. Use the perimeter equation to solve for either x or d, and plug this expression into the area equation. Take the derivative, set it to zero, and solve for either x or d, which ever you used in your area equation. Then solve for the other variable using the perimeter equation, and solve for the area. I get max area of the rectangle being 20,000/pi.

As far as the second question goes, take the derivative of the expression, set it to zero, and solve for x. If x occurs on the interval [2,3], then you have a stationary point and you can look at the second derivative to determine if it is a maximum, a minimum, or a saddle point. If the derivative is non-zero over the whole interval, then one of the two endpoints is the maximum.

R

 

[Shooters]

Originally posted by: PHiuR
hey, thanks for replying.

why would you use the width as 0.5?

Well, you know that the total perimeter of the track is 400. Think of the perimeter as being made up of 4 sections, two straight and two semicircular. The perimeter of the two semicircular sections will be the circumference of one whole circle, which is 2*pi*r like you stated.

Therefore, the total length of the two remaining straight sections will be the total perimeter (which is 400) minus the perimeter of the two semicircles (which is one whole circle, 2*pi*r).

However, the width of the rectangle is the length of only one of these straight sections, hence you have to mutliply by 0.5.

So, that gives width = 0.5*(400 - 2*pi*r)

 

[PHiuR]

Originally posted by: rgwalt
I get the same answer as Shooters for the area of the rectangle. Area of the rectangle is A=x*d where d is the diameter of the circles. Perimeter of the track is 400 = pi*d+2x. Use the perimeter equation to solve for either x or d, and plug this expression into the area equation. Take the derivative, set it to zero, and solve for either x or d, which ever you used in your area equation. Then solve for the other variable using the perimeter equation, and solve for the area. I get max area of the rectangle being 20,000/pi.

As far as the second question goes, take the derivative of the expression, set it to zero, and solve for x. If x occurs on the interval [2,3], then you have a stationary point and you can look at the second derivative to determine if it is a maximum, a minimum, or a saddle point. If the derivative is non-zero over the whole interval, then one of the two endpoints is the maximum.

R

can anyone show how to do question 2? argh i hate calc and love sleep...love:hate relation haha... oh well

 

[PHiuR]

Originally posted by: Shooters

Originally posted by: PHiuR
hey, thanks for replying.

why would you use the width as 0.5?

Well, you know that the total perimeter of the track is 400. Think of the perimeter as being made up of 4 sections, two straight and two semicircular. The perimeter of the two semicircular sections will be the circumference of one whole circle, which is 2*pi*r like you stated.

Therefore, the total length of the two remaining straight sections will be the total perimeter (which is 400) minus the perimeter of the two semicircles (which is one whole circle, 2*pi*r).

However, the width of the rectangle is the length of only one of these straight sections, hence you have to mutliply by 0.5.

So, that gives width = 0.5*(400 - 2*pi*r)

i tried useing rgwalts way, and i got too

63.662/d * d


for first deriv, but i dont kno where to go now..anyone? (kinda early to be doing this kinda stuff haha.

 

[PHiuR]

argh i solved it and got 6266.2 as the maxx area

 

[rgwalt]

Originally posted by: PHiuR
argh i solved it and got 6266.2 as the maxx area

I get 6366.2 as an answer, so you are off by 100 or typed a number wrong.

We really need a homework forum here...

R

 

[gflores]

Originally posted by: rgwalt

Originally posted by: PHiuR
argh i solved it and got 6266.2 as the maxx area

I get 6366.2 as an answer, so you are off by 100 or typed a number wrong.

We really need a homework forum here...

R

Here it is...