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Difference between quadratic and linear Watch

1. I have a quadratic and a linear equation. Both are plotted and the line intersects at 2 points. I now need to find where the difference between the line and quadratic is greatest.. how can I do this?
2. (Original post by Carlo08)
I have a quadratic and a linear equation. Both are plotted and the line intersects at 2 points. I now need to find where the difference between the line and quadratic is greatest.. how can I do this?
Differentiate?

Simultaneousness equations?
3. (Original post by Student#254)
Differentiate?

Simultaneousness equations?
No.

I thought maybe the curve which goes under the line may be symmetric, in which case the greatest difference would lie in the middle ... and having tested that it seems to work ... would this be right?
4. (Original post by Carlo08)
No.

I thought maybe the curve which goes under the line may be symmetric, in which case the greatest difference would lie in the middle ... and having tested that it seems to work ... would this be right?
Do you mean that you would complete the square on the quadratic and find the vertex?

I'm not sure to be honest. Maybe someone else will help you out
5. (Original post by Carlo08)
No.

I thought maybe the curve which goes under the line may be symmetric, in which case the greatest difference would lie in the middle ... and having tested that it seems to work ... would this be right?
This isn't true though. All you need to do is consider the gradients. If the curve is below for example, and the line is upward sloping then the distance will keep increasing until the gradient of the curve becomes positive and is the same as the line. Then as x increases further, the distance will become smaller because the gradient of the curve is greater than that of the line, and hence eventually cuts it.
6. (Original post by Carlo08)
I thought maybe the curve which goes under the line may be symmetric, in which case the greatest difference would lie in the middle ... and having tested that it seems to work ... would this be right?
Not necessarily, consider if the line's point of intersection is at the minima. In this case, the smallest distance is at the middle!
(Edit: ninja'd)

It depends on how you're defining the distance here, do you mean at particular x value, the vertical difference between the two lines?

Let's say our quadratic is Q(x), and our linear "line" is L(x).

The distance between them at any point is | Q(x) - L(x) |.

So all we have to do is find the maximum of this new function, which we can do, right?

PS. You may want to rephrase the question, I'm fairly sure if you pick a big enough x value then the distance between Q(x) and L(x) can be as big as you like. Do you mean the maximum distance we can find for x in between the two points of intersection?
7. (Original post by StephenNeill)
Not necessarily, consider if the line's point of intersection is at the minima. In this case, the smallest distance is at the middle!
(Edit: ninja'd)

It depends on how you're defining the distance here, do you mean at particular x value, the vertical difference between the two lines?

Let's say our quadratic is Q(x), and our linear "line" is L(x).

The distance between them at any point is | Q(x) - L(x) |.

So all we have to do is find the maximum of this new function, which we can do, right?

PS. You may want to rephrase the question, I'm fairly sure if you pick a big enough x value then the distance between Q(x) and L(x) can be as big as you like. Do you mean the maximum distance we can find for x in between the two points of intersection?
Hey, Ok , we have 2 lines. One a curve from the equation 5x^2+700x+5280 gives the total cost of production

A straight line given by 1200x gives the revenue

I have found where these two lines insersect by equating the equations which gives the points 12 and 88.

So the curve slices the y axis at 5280 and continues to curve towards infinity with the line cutting in at 12 and 88.

I need to find the value of the biggest profit which is where the difference between revenue and cost is greatest .. ie where the curve lies furthers below the line.

I figure seeing as its a curve of x^2 that it should be symetric between any 2 points no? in which case the biggest difference would be at (88-12)+12 and then plug this value into the quadratic.

If i find the minimum of the curve it will give me the global minimum which is way off into the negative side of the axis.
8. (Original post by Carlo08)
Hey, Ok , we have 2 lines. One a curve from the equation 5x^2+700x+5280 gives the total cost of production

A straight line given by 1200x gives the revenue

I have found where these two lines insersect by equating the equations which gives the points 12 and 88.

So the curve slices the y axis at 5280 and continues to curve towards infinity with the line cutting in at 12 and 88.

I need to find the value of the biggest profit which is where the difference between revenue and cost is greatest .. ie where the curve lies furthers below the line.

I figure seeing as its a curve of x^2 that it should be symetric between any 2 points no? in which case the biggest difference would be at (88-12)+12 and then plug this value into the quadratic.

If i find the minimum of the curve it will give me the global minimum which is way off into the negative side of the axis.
A much easier way to do it is to say profit= TR-TC and maximise that...
9. (Original post by little_wizard123)
A much easier way to do it is to say profit= TR-TC and maximise that...
This is how I found the intersections.

Break even is TR-TC=0 Find roots of new quadratic.

This isnt actually for me but for a friend. She has never learnt calculus so I figure there must be an easy way.
10. (Original post by Carlo08)
This is how I found the intersections.

Break even is TR-TC=0 Find roots of new quadratic.

This isnt actually for me but for a friend. She has never learnt calculus so I figure there must be an easy way.
Yeah, solving TR-TC=0 gives you the intersections, and d(TR-TC)/d(x) where x is the quantity will give you the maximum profit. i.e. just maximise the difference between revenue and costs.
11. (Original post by little_wizard123)
Yeah, solving TR-TC=0 gives you the intersections, and d(TR-TC)/d(x) where x is the quantity will give you the maximum profit. i.e. just maximise the difference between revenue and costs.
This doesn't seem to work.

When plotted on excel I can see the maximum difference occurs at x=50

By doing dtr-tc/dx I get 25x-500

so setting to 0 gives x as 20. :s
12. (Original post by Carlo08)
This doesn't seem to work.

When plotted on excel I can see the maximum difference occurs at x=50

By doing dtr-tc/dx I get 25x-500

so setting to 0 gives x as 20. :s
How did you get that for your differentiation?
13. d(TR-TC)/dx = 0

=> 1200 - (10x - 700) = 0.

10x = 500 giving x=50.

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