Maximum area between two curves within a given interval

[SithsNGiggles]

Homework Statement

Consider the area bounded between the curves $y=3-x^2$ and $y=-2x$. Suppose two vertical lines, one unit apart, intersect the given area. Where should these lines be placed so that they contain a maximum amount of the given area between them? What is this maximum area?

Homework Equations

The Attempt at a Solution

First I computed the area between the two curves using the integral
$A = \int_{-1}^{3} (3 - x^2) - (-2x) dx$
$A = \int_{-1}^{3} 3 + 2x - x^2 dx$
$A = \frac{32}{3}$.

Then I set up another integral for the maximum area between two vertical lines $y = k$ and $y = k+1$:
$A(k) = \int_{k}^{k+1} 3 + 2x - x^2 dx$, for $k \in [-1, 3]$.

I'm not sure where to go from here, but something tells me there's some application of the first derivative test that I can put to use here. Unfortunately I'm not sure how to apply it. Any help is appreciated, thanks.

 

[LCKurtz]

Homework Statement

Consider the area bounded between the curves $y=3-x^2$ and $y=-2x$. Suppose two vertical lines, one unit apart, intersect the given area. Where should these lines be placed so that they contain a maximum amount of the given area between them? What is this maximum area?

Homework Equations

The Attempt at a Solution

First I computed the area between the two curves using the integral
$A = \int_{-1}^{3} (3 - x^2) - (-2x) dx$
$A = \int_{-1}^{3} 3 + 2x - x^2 dx$
$A = \frac{32}{3}$.

Then I set up another integral for the maximum area between two vertical lines $y = k$ and $y = k+1$:
$A(k) = \int_{k}^{k+1} 3 + 2x - x^2 dx$, for $k \in [-1, 3]$.

I'm not sure where to go from here, but something tells me there's some application of the first derivative test that I can put to use here. Unfortunately I'm not sure how to apply it. Any help is appreciated, thanks.

You need to find critical points of A(k), so you need its derivative. You can either integrate to get the formula then differentiate that or use the Fundamental Theorem of calculus to get the derivative A'(k). Either way set the derivative equal to zero to look for a max.

 

[Alpha Floor]

First of all, be careful with how you define the domain for "k". You sure the upper limit is 3? Think that the lines are placed at k and k+1, so that the second line should also fall into the area bounded by both curves.

You didn't really need to calculate the area bounded by the curves, that was not asked.

You seem to know how to calculate the area between the curves and the vertical lines as a function of your parameter "k". That area, called A(k), is a function which you want to maximize, doesn't that sound familiar? You derive A(k) and find the roots of dA/dk = 0 for k inside the valid domain... that's it. Finally calculating the maximum area shouldn't be diffcult!

 

[SithsNGiggles]

Ah, I see how to go about this now. I wasn't sure if the integral over [k, k+1] was a function of x or k.

To address the concerns brought up: I see why you say I should set up the restriction $k \in [-1, 2]$ to account for $k + 1$. At the time, I thought it wouldn't matter. My reasoning was that, with the way the integral is set up, any area beyond the line x = 3 will be negative, so the area on the interval [k, k + 1] in such a scenario cannot be the maximum area. (I hope that made sense.)

Finding out the area between the two curves was a previous question, and I thought I might have needed it for this part so I included it, too.

$A'(k) = \frac{d}{dk}[ [3x + x^2 - \frac{1}{3}x^3]_{k}^{k+1}]$
$A'(k) = \frac{d}{dk}[ (3(k + 1) + (k + 1)^2 - \frac{1}{3}(k+1)^3) - (3k+ k^2 - \frac{1}{3}k^3)]$
$A'(k) = \frac{d}{dk}[ \frac{11}{3} + k - k^2 ]$
$A'(k) = 1 - 2k$

And from here I'll carry on with the derivative test. Thanks for the help!