# Homework Help: Equal area divided by line

1. Apr 28, 2016

### terryds

1. The problem statement, all variables and given/known data

In the picture above, line y = c intersects with parabola y = 6x-x^2 in the first quadrant.
If the gray area below line y = c and the gray area above line y=c are equal, then value of c is ...
A. 19/4
B.21/4
C.23/4
D.25/4
E. 27/4

2. Relevant equations

Area under parabola = 2/3* base * height
Area above parabola = 1/3 * base* height
Area = definite integral

3. The attempt at a solution

Parabola y=6x-x^2 has peak point (3,9)

Suppose a is the x-intercept of the line and parabola.
At x = a, c equals 6a-a^2
So, the gray area below the line is 1/3 * a * (6a-a^2)

The base for the upper gray parabola area is 2(3) - 2a = 6-2a
So, the gray area above the line is 2/3 * (6-2a) * (9-(6a-a^2))

I get a = 1.592
and c = 6a-a^2 = 7.017

However, I'm sure the correct answer is 27/4 (I've tested it using integral method)
Please help...

2. Apr 28, 2016

### Ray Vickson

Your expression for the gray area below the line is incorrect.

Last edited: Apr 28, 2016
3. Apr 28, 2016

### blue_leaf77

Translate the parabola downward till the line y=c coincides with the x axis and then integrate the translated parabola from x=0 to the right intersection with the x axis. You will find c by making this integral vanish.

4. Apr 28, 2016

### blue_leaf77

Alternatively you can work with the inverse function $x=f^{-1}(y)$. I think this way can br easier.

5. Apr 29, 2016

### terryds

Why??
I think it's correct

Okay, so the translated parabola become

y = -x^2+6x-c

But, I don't know the roots of the polynomial since there is an unknown c.
Even if I substitute it with ( -6 ± 36 - 4k ) / -2 , I see it'll be very messy and hard to solve

The inverse function of the parabola is

f^-1(x) = √(9-x) + 3

But, what does the inverse have to do with the area ? I don't understand the method

6. Apr 29, 2016

### blue_leaf77

I have tried following the method suggested in post #4. I ended up having to solve cubic polynomial in $c$. However using graphical method I did find 27/4 as one of the solutions. Probably there are other members who can propose a more elegant method.

7. Apr 29, 2016

### terryds

Okay..
Thanks anyway

8. Apr 29, 2016

### ehild

Anyway, you can write the solutions x1 and x2 (x1<x2) of the quadratic equation in terms of c. $\int _0 ^{x_1}(c-y)dx = \int _{x_1} ^{x_2}(y-c)dx$ means that $\int _0 ^ {x_2} {(-x^2+6x-c)dx}=0$
Do the integration, substitute the limits, and solve the equation for c (it will become a quadratic one), not that terrible.

9. Apr 29, 2016

### blue_leaf77

I haven't tried myself what I suggested in post #3 which seems to coincide to your proposed method, but is it really quadratic in $c$?
I mean first we need to solve for $x_2$, probably by using the usual quadratic root formula, and this will sure contain $c$ under a square root. As for the integral itself, there will be $x^3$ arising from the quadratic term in $y(x)$.

10. Apr 29, 2016

### ehild

Yes, you proposed the same method.
It will be simplified to a quadratic, as all terms of the definite integral contain x2.

11. Apr 29, 2016

### Ray Vickson

12. Apr 29, 2016

### terryds

Thanks.. I've got it

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