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F22b + MP1A = mad fun for me now. . .
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Same basic principle as the other problem.
I'm not going to do it for you this time. I'll get you started though
The two top points of the rectangle are bound by the parabola, so your equation for the area of the rectangle should have something to do with that equation. From there, you need to figure out how to find the maximum of that rectangle, just as in the problem before.
It's going to be a squre, just like in the previous problem. Just remember, the maximum area of any rectangle with fixed boundaries (be it perimeter, width, height, or bound by a function like this example) is a square. However, don't let that be your entire answer, you need to figure out how to do the math behind it.
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Let's find the total area of the parabola using the ellipse equation and then dividing that by 2. This will give us the total area under the parabola. The equation is of an ellipse is (x^2/a^2 + y^2/b^2) = 1, where a and b are the foci of the ellipse. That is the points that touch on the axis or the maximum curving point, also referred to as an vertex on a parabola or when the slope is zero in the x-axis direction or undefined in the y-axis direction. Find a and b. You already have them actually. What are the intercepts of the parabola again on the x and y axis? This will be your a and b.The Lord watches over me!
"Stop punching down on my people!!!"
- D. Chappelle
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Originally posted by Tnwagn View PostSame basic principle as the other problem.
I'm not going to do it for you this time. I'll get you started though
The two top points of the rectangle are bound by the parabola, so your equation for the area of the rectangle should have something to do with that equation. From there, you need to figure out how to find the maximum of that rectangle, just as in the problem before.
It's going to be a squre, just like in the previous problem. Just remember, the maximum area of any rectangle with fixed boundaries (be it perimeter, width, height, or bound by a function like this example) is a square. However, don't let that be your entire answer, you need to figure out how to do the math behind it.F22b + MP1A = mad fun for me now. . .
MY RIDE
My swap parts list and pricing
vouches:
Bought from: Smeagren83(quite a few times now LOL), 97lude, AZaccord, M3torz2nR, g7kobayashi, sickoffthe206 and Kurobei, Jarhead, prNonVtec4u, caserX
Sold to: Snailin91, Smeagren83, cb7dazz, Drummersteve7, Slick
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Originally posted by DA92CB7 View Postthe co-ordinates where the parabola cross the y-axis are -3.162278,0 an 3.162278,0 since the parabolas cut in half by the y axis these points are reciprocals of each other so there was no need to find both just one.F22b + MP1A = mad fun for me now. . .
MY RIDE
My swap parts list and pricing
vouches:
Bought from: Smeagren83(quite a few times now LOL), 97lude, AZaccord, M3torz2nR, g7kobayashi, sickoffthe206 and Kurobei, Jarhead, prNonVtec4u, caserX
Sold to: Snailin91, Smeagren83, cb7dazz, Drummersteve7, Slick
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Right. I just want to make sure you understand. The maximum is going to be a rectangle which is a square. If you find the maximum area under the parabola, you can set up inequalities to maximize the L and W and get the maximum for the area of the window. That's the way I would do it if not using calculus.The Lord watches over me!
"Stop punching down on my people!!!"
- D. Chappelle
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so since we are dealing with actual lengths it would be the positive one right? But how do I put it into an equation where I can use substitution. I think I need to do this the algabraic way so I am not getting too far ahead of myself.F22b + MP1A = mad fun for me now. . .
MY RIDE
My swap parts list and pricing
vouches:
Bought from: Smeagren83(quite a few times now LOL), 97lude, AZaccord, M3torz2nR, g7kobayashi, sickoffthe206 and Kurobei, Jarhead, prNonVtec4u, caserX
Sold to: Snailin91, Smeagren83, cb7dazz, Drummersteve7, Slick
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Originally posted by DA92CB7 View Postso since we are dealing with actual lengths it would be the positive one right? But how do I put it into an equation where I can use substitution. I think I need to do this the algabraic way so I am not getting too far ahead of myself.
The area of any rectangle square is L x H. We then have a formula of a parabola which is in the form of y = ax^2 + bx + c. The y is the H and the x is the L. Also, the area of a parabola is (2/3) L x W.
We have:
A, parabola = (2/3) L x W => y = 10 - x^2
A, square = L x W
Now we have to use some of this info to find the maximum square that can fit into the parabola.The Lord watches over me!
"Stop punching down on my people!!!"
- D. Chappelle
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I came up with the top corner of the square being at the point x, 10-x^2 I just substituted y for its equivalant so the new equation would be 2x*(10-x^2) right?F22b + MP1A = mad fun for me now. . .
MY RIDE
My swap parts list and pricing
vouches:
Bought from: Smeagren83(quite a few times now LOL), 97lude, AZaccord, M3torz2nR, g7kobayashi, sickoffthe206 and Kurobei, Jarhead, prNonVtec4u, caserX
Sold to: Snailin91, Smeagren83, cb7dazz, Drummersteve7, Slick
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Originally posted by DA92CB7 View PostI came up with the top corner of the square being at the point x, 10-x^2 I just substituted y for its equivalant so the new equation would be 2x*(10-x^2) right?The Lord watches over me!
"Stop punching down on my people!!!"
- D. Chappelle
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zero or y again cause they basically mean the same thing.F22b + MP1A = mad fun for me now. . .
MY RIDE
My swap parts list and pricing
vouches:
Bought from: Smeagren83(quite a few times now LOL), 97lude, AZaccord, M3torz2nR, g7kobayashi, sickoffthe206 and Kurobei, Jarhead, prNonVtec4u, caserX
Sold to: Snailin91, Smeagren83, cb7dazz, Drummersteve7, Slick
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