More HW help . . .

[DA92CB7]

yup

[Straight Success]

Alright let me take a quick #1, and I'll be back in 2 minutes. I just had two glasses of hot tea.

[ibr_adam09]

What do you mean?

Success + homework = perfect.

[Tnwagn]

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.

[Straight Success]

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.

[Straight Success]

Once you have that the formula for the area of a parabola is (2/3) B x H or (2/3) L x W. So us a and b as the L and W, and find the area of the parabola.

[DA92CB7]

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.

yeah I noticed it would be the same an I figured it was a square because of the maximums you were talking about in the last one. I am not looking for someone to complete this for me I just couldn't figure out how to go through the steps. The other problem I was doing my math wrong I knew that once I read the link that was posted thats what I needed just fucked up the number crunching.

[DA92CB7]

the 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.

^^^^

[Straight Success]

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.

[DA92CB7]

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.

[Straight Success]

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.

Okay. Where going to use the substitution method. My bad for everything else. I'm use to helping people solve these problems taking the shortest possible route to save on time.

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.

[Straight Success]

What did you get for the area of the parabola?

[DA92CB7]

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?

[Straight Success]

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?

...is equal to... you left that part out.

[DA92CB7]

zero or y again cause they basically mean the same thing.