optimisation problem: Maximize area of rect above x-axis, under y=12-x^2

[Robot]

Hello all,

I was not able to put up a new thread so I am asking my question in an old thread.
Apologies for that.

Maximise the area of the rectangle inscribed under the parabola above x-axis in first quadrant? y=12-x^2

I know how to solve the problem.

My question is Can I take the area of the whole rectangle (both in FIRST AND SECOND QUADRANT ) as my main function ?
But based on domain negative value of x will be discarded 0<x<sqrt 12.

You get the same answer even if you consider the area only in first quadrant.
I want to know if the main function can be the area of the whole rectangle (first and second quadrant ) or not ?
Thank you

 

[mmm4444bot]

Maximise the area of [a] rectangle inscribed under the parabola [in the] first quadrant?

y = 12 - x^2

My question is Can I take the area of the whole rectangle (both in FIRST AND SECOND QUADRANT ) as my main function ?

Well, you can create any kind of function you like, as long as it leads to the correct answer.

However, the exercise asks for the maximum area of a rectangle in Quadrant I. You may read it as "under the parabola in the first quadrant only". Therefore, you need to report the area of the largest rectangle possible, in the first quadrant as described.

You get the same answer even if you consider the area only in first quadrant.

I must be misunderstanding you, here. If you report the area of the largest inscribed rectangle spanning both Quadrants I & II, it will have twice the area (due to symmetry about the vertical axis) as the largest rectangle in Quadrant I only.

What do you get for your answer?

 

[Robot]

rectangle in both quadrants

Lets say I create the function spanning both quadrants. If I half it at the end, will that be correct.

Actually I am a TA. I feel like this is wrong because the correct function is the area of the rectangle only in first quadrant.
What do you think ?

Well, you can create any kind of function you like, as long as it leads to the correct answer.

However, the exercise asks for the maximum area of a rectangle in Quadrant I. You may read it as "under the parabola in the first quadrant only". Therefore, you need to report the area of the largest rectangle possible, in the first quadrant as described.

I must be misunderstanding you, here. If you report the area of the largest inscribed rectangle spanning both Quadrants I & II, it will have twice the area (due to symmetry about the vertical axis) as the largest rectangle in Quadrant I only.

What do you get for your answer?

 

[stapel]

Maximise the area of the rectangle inscribed under the parabola y = 12 - x^2 and above the x-axis in first quadrant.

I know how to solve the problem.

My question is Can I take the area of the whole rectangle (both in FIRST AND SECOND QUADRANT ) as my main function ?

What do you mean by "as my main function"?

But based on domain negative value of x will be discarded 0<x<sqrt 12.

"Based on domain" how? Was there additional information for this exercise, not yet posted to this thread, which said that the rectangle had to be above the x-axis (so positive y-values) below the given parabola, and also to the right of the y-axis (so positive x-values)?

If so, please reply with all of the missing information. If not, on what basis are you "discarding" the included negative x-values for this exercise?

You get the same answer even if you consider the area only in first quadrant.

Who is "you"? How does one "get the same answer even if you consider" only half of the (currently stated) domain?

You say that you "know how to solve the problem". Please reply showing your solution, so we can see what you're talking about. Thank you!

 

[JeffM]

I think what is being asked is whether, to compute the maximum area of a rectangle bounded by the x-axis, y-axis, and the given parabola, you can halve the maximum area of a rectangle bounded by the x-axis, and the given parabola. As mmm has already said, symmetry shows that you can.

\(\displaystyle area\ in\ first\ quadrant = \dfrac{1}{2} * area\ in\ first\ and\ second\ quadrants.\)

 

[JeffM]

\(\displaystyle \text {Area of rectangle in first quadrant} = a_1.\)

\(\displaystyle \text {Area of rectangle in first and second quadrant} = a_2.\)

\(\displaystyle a_1 = (x - 0) * (12 - x^2) = 12x - x^3 \implies\)

\(\displaystyle \dfrac{da_1}{dx} = 12 - 3x^2 \implies\)

\(\displaystyle \dfrac{da_1}{dx} = 0 \implies x^2 = \dfrac{12}{3} \implies x = 2 \text { and } a_1 = 12 * 2 - 2^3 = 16.\)

\(\displaystyle a_2 = \{x - (-\ x)\}(12 - x^2) = 2x(12 - x^2) = 24x - 2x^3 \implies\)

\(\displaystyle \dfrac{da_2}{dx} = 24 - 6x^2 \implies\)

\(\displaystyle \dfrac{da_2}{dx} = 0 \implies x^2 = \dfrac{24}{6} \implies x = 2 \text { and } a_2 = 24 * 2 - 2 * 2^3 = 32.\)

In short, \(\displaystyle a_1 = \dfrac{1}{2} * a_2.\)

Of course, without symmetry, the process would not work so it is generally wiser to follow the problem's instructions exactly, but the process is completely valid here because of symmetry.

 

[mmm4444bot]

Lets say I create the function spanning both quadrants. If I half it at the end, will that be correct.

Yes.

In Quadrants I & II, the base of the rectangle is 2·x and the height is f(x).

answer = [(2·x)·f(x)]/2 using the appropriate value of x.

In Quadrant I only, the base is x and the height is f(x).

answer = x·f(x) using the appropriate value of x.

These area expressions are exactly the same, after simplifying the first by cancelling those factors of 2. :cool: