Help Solve Optimization Problem

[JenniferBlanco]

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

Homework Statement

Homework Equations

y=x^2

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

The Attempt at a Solution

I don't know where to start :(

 

[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.

 

[JenniferBlanco]

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.

How do I calculate the area of the segment?

 

[Rainbow Child]

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

 

[JenniferBlanco]

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

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?

 

[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?

 

[JenniferBlanco]

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?

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

 

[Rainbow Child]

Correct! But it is a definite integral, so

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

 

[JenniferBlanco]

now how shall I proceed?

 

[Rainbow Child]

Did you find I?

 

[JenniferBlanco]

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

 

[Rainbow Child]

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

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

 

[JenniferBlanco]

So that just becomes

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

?

 

[Rainbow Child]

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

 

[JenniferBlanco]

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

Ooops! let me 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

 

[Rainbow Child]

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

 

[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

 

[Rainbow Child]

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

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 $$x=-1, \,x=3$$.

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?

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!

 

[JenniferBlanco]

Thanks for the explanation about the definite integral

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!

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

 

[Rainbow Child]

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

 

[JenniferBlanco]

yes it does

 

[Rainbow Child]

Ok! How can you calculate an area of a triangle if you know the coordinates of the vertices?

 

[JenniferBlanco]

but do we don't know the coordinates of point P.

1/2 * 8.944 * height

 

[Rainbow Child]

Yes, but we know the propety that P satisfies.

 

[JenniferBlanco]

Yes, but we know the propety that P satisfies.

hmm so how do we go about using the the maximum distance from QR statement?

 

[Rainbow Child]

That's the easy part

We already now the equation of the line QR. Let the coordinates of P be $$(x_0,x_0^2)$$. Now what's the distance of P from the line QR?

 

[JenniferBlanco]

The distance equation is Ax+By+C/ sqrt (a^2+b^2)

so the distance will be 2(x0) + 1(xo^2) +3 / sqrt(4+1)

 

[Rainbow Child]

You need an absolute value, and I think that

$$d=\frac{|2x_0-x_0^2+3|}{\sqrt{5}}$$

Correct?

 

[JenniferBlanco]

yes, you're right

 

[Rainbow Child]

Good!

Now what's the maximum value of $$d$$? For which $$x_0$$ you obtain that?

 

[JenniferBlanco]

Good!

Now what's the maximum value of $$d$$? For which $$x_0$$ you obtain that?

The max value I get is when $$x_0=1$$ and D would then be 4/sqrt(5)

 

[Rainbow Child]

Very nice! Let us now be proud of ourselves and calculate the area of the triangle!

 

[JenniferBlanco]

Very nice! Let us now be proud of ourselves and calculate the area of the triangle!

So area of the triangle = 1/2 * 8.944 *4 = 17.88 units^2

I got the midpoint of QR as (1,5) and the distance from there was 4.

So my final answer for the area of the triangle is 17.88 units^2

 

[Rainbow Child]

No, the distance is $$d=\frac{4}{\sqrt{5}}$$=height. You don't need the midpoint

 

[JenniferBlanco]

ohhh. then the Area = 1/2 * 8.944 * 4/sqrt(5) = 7.999 which is = 8units^s [our earlier answer ]