Optimization problem

1. Jan 22, 2008

JenniferBlanco

I'm having a tough time solving this question. I'd appreciate it if someone can please help me out

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

2. Relevant equations

y=x^2

Triangle QPR =3/4 of the area of the parabolic segment enclosed between QR and the parabola

3. The attempt at a solution

I don't know where to start :(

2. Jan 22, 2008

NateTG

Well, the area of the segment should be pretty easy to work out. Why don't you start there? It might give you some ideas on how to calculate the area of the triangle.

3. Jan 22, 2008

JenniferBlanco

How do I calculate the area of the segment?

4. Jan 22, 2008

Rainbow Child

Do you know how to calculate the area between two functions $$f(x),g(x)$$ for $$x \in (a,b)$$?

5. Jan 22, 2008

JenniferBlanco

Nope, we haven't been taught that. We actually just started with integration today. Anyways, I think I kind of got the first step though -

y=x^2

and then the equation of the line b/w Q and R can be found out
m = 9-1/ 3-(-1) = 8/4= 2
y-1=2(x+1)
y=2x+3

elminating y and we get:
x^2 - 2x -3

now what shall I do?

6. Jan 22, 2008

Rainbow Child

The area between two functions $$f(x),g(x)$$ for $$x \in (a,b)$$ is

$$\int_a^b|f(x)-g(x)|\,d\,x$$

In your case $$f(x)=2x+3,\,g(x)=x^2$$.

Now can you caclulate the area of the segment?

7. Jan 22, 2008

JenniferBlanco

We just started integrating today, So I am not that confident about it :( I'll give it a shot though

2x+3-x^2
[(2x^2)/2] +3x- [(x^3)/3]
[x^2]+3x-[(x^3)/3]

I don't know what to do now

8. Jan 22, 2008

Rainbow Child

Correct! But it is a definite integral, so

$$I=\int_a^b|f(x)-g(x)|\,d\,x=(x^2+3x-\frac{x^3}{3})\Big|_{-1}^{3}$$

9. Jan 22, 2008

JenniferBlanco

now how shall I proceed?

10. Jan 22, 2008

Rainbow Child

Did you find I?

11. Jan 22, 2008

JenniferBlanco

I don't know how to do that. Do I have to plug -1 and 3 into x?

12. Jan 22, 2008

Rainbow Child

Ok!
$$(x^2+3x-\frac{x^3}{3})\Big|_{-1}^{3}$$ means

$$(x^2+3x-\frac{x^3}{3})\Big|_{x=3}-(x^2+3x-\frac{x^3}{3})\Big|_{x=-1}$$

13. Jan 22, 2008

JenniferBlanco

So that just becomes

9 -(2 - (-1/3))
9 -(2.33)
6.67 units^2

?

14. Jan 22, 2008

Rainbow Child

I think its better $$\frac{32}{3}$$

15. Jan 22, 2008

JenniferBlanco

Ooops! lemme recheck
we first plug in 3 and get

[9+9 - (27/3)]
=9

and for the other side we get

-[1-3 - [-1/3] ]
-[-2 + [1/3]]
-[-1.67]
1.67

9+1.67 = 10.67 or 32/3 :)

aah so the area of the parabolic segment = 32/3 units^2
while the area of the Triangle = (3/4)*(32/3) = 96/12 = 8 units^2

16. Jan 22, 2008

Rainbow Child

Yes! But I think you have to calulate the area of the triangle independently.

17. Jan 22, 2008

JenniferBlanco

Thank you!

But doesn't it say that "Triangle QPR =3/4 of the area of the parabolic segment enclosed between QR and the parabola"

so that makes the triangle areas 3/4 of 32/3units^2 which is equal to 8..right?

also can you please explain how you got 3 and -1 for the numbers to be plugged in?

Thanks
Jennie

18. Jan 22, 2008

Rainbow Child

The geometric meaning of the definite integral

$$\int_a^b|f(x)| \,d\,x$$

is the area between the graph of $$f(x)$$ the x-axis and the vertical lines $$x=a,x=b$$

When you have two curves and you want the area between them you should use

$$\int_a^b|f(x)-g(x)| \,d\,x$$

where now $$x=a$$ and $$x=b$$ are the points where the two urves intersect. In your case these are $$x=-1, \,x=3$$.

From the statement of the problem I think that first you should calculate the area of the triangle too, in order to see that Arcimedes was correct!

But I don't I take an oath! :rofl:

19. Jan 22, 2008

JenniferBlanco

Thanks for the explanation about the definite integral

Now how do we go about calculating the area of the triangle?

20. Jan 22, 2008

Rainbow Child

I don't understand what $$\overline{QR}$$ stands for. Does it stands for the chord QR?