More HW help . . .

[DA92CB7]

no wait im dumb that represents the area of the "rectangle"

[Straight Success]

This is tedious. This is how I did it.

Calculus way:

Found the area under the parabola by integrating the function from 0 to 10^(1/2) and multiplying that by 2. I got 42.16.

Then I set up inequalities.

0 <= L <= 2*10^(1/2) ............................... Min. and Max. values for the length.
0 <= H <= 10 ....................................Min. and Max. values for the height.
0 <= LH < 42.16 .....................................Min. and Max. values for the entire window.

Solving it out for L and H using substitution, I got the maximum area is 24.3.

Geometric way

Using the area of a parabola formula, I found the total area of the parabola to be 42.16.

Then I used the same inequalities.

[vietkid_2006]

my head hurts seeing all of this math...

[DA92CB7]

This is tedious. This is how I did it.

Calculus way:

Found the area under the parabola by integrating the function from 0 to 10^(1/2) and multiplying that by 2. I got 42.16.

Then I set up inequalities.

0 <= L <= 2*10^(1/2) ............................... Min. and Max. values for the length.
0 <= H <= 10 ....................................Min. and Max. values for the height.
0 <= LH < 42.16 .....................................Min. and Max. values for the entire window.

Solving it out for L and H using substitution, I got the maximum area is 24.3.

Geometric way

Using the area of a parabola formula, I found the total area of the parabola to be 42.16.

Then I used the same inequalities.

so the calc way would be 10*2*sqrt(10) and you should come out with the 42.16?

[DA92CB7]

its max hight cannot be 10 because it is a square inside a porabola that maxes out at 10 so the hight is something x - 10 or 10-x^2 but I don't know what else to do after that.

and in the original equation all I needed to find was the highth and width of the "rectangle" not its area

[Straight Success]

so the calc way would be 10*2*sqrt(10) and you should come out with the 42.16?

NO. The calculus way is like this...

Twice the integral of 10 - x^2 dx, integrated from 0 to Root 10 is 42.16.

Don't confuse yourself. Just use the area formula for the parabola.

A = (2/3)*L*W = (2/3)*10*2[(10)^(1/2)] = 42.16.

10^(1/2) is the same as Root 10, which is the x coordinate.
The two in front of the Root 10 id both sides of the parabola, from - Root 10 to + root 10.

[Straight Success]

Just call me. I can explain it easier over the phone. (267) 438-0457.

[Straight Success]

All done. It's right.

[DA92CB7]

when I solved this equation with a girl I had her hella confused, but there was an easier way of doing it as far as what I know but I can't remember it again HAHAHAHAHA thanks for the help straight!

[Straight Success]

when I solved this equation with a girl I had her hella confused, but there was an easier way of doing it as far as what I know but I can't remember it again HAHAHAHAHA thanks for the help straight!

Cool. It was rather quick for me. I did it and the calculations in about 1 min. and 45 secs. Probably because I understand the mathematics well. Good luck. By the way, was it correct?

[apuforyou]

This is what I get for sleeping in math classes when I was in highschool, I didn't understand a single word any of you said! Might was well be a Korean talking extremely fast in German with a Spanish accent

I'm a failure as an asian...


(ok ok, I'm only half filipino)