Question

how do we maximise area here

Answer 1

You have to start with finding two functions that have variables include both the area of a triangle and the area of the curves y=ax+b and y=x^2. However, the find the area of those two curves you need to set up an integration. Have you learned any integration techniques?

Answer 2

Wrong, you don't need to know any integration to solve this problem Msdoroff.

Essentially what you are doing here is finding an equation for the area of that triangle. When you take the derivative you're really finding the point where that area formula is the highest since that will be the peak of the curve.

So how do you find the area of the triangle? Remember the formula from geometry is A=(1/2)bh. So we just need to find the length of the base and the height in terms of these two formulae. Our base and height are really interchangeable here and are the lines AP and BP.

To find the points A and B we know that they are the points where y=mx+b intersects y=x^2. So we simply set the equations equal to each other and put them into the quadratic formula and get:\[x _{A}=(m+\sqrt{m^2+4b})/2\]\[x _{B}=(m-\sqrt{m^2+4b})/2\]
And then you can either plug that x-value into y=x^2 or y=mx+b to find your y-value.

Use these equations for x and y coordinates to find a general equation by using the distance formula for the lines AP and AB, remember, P is any point along y=x^2 between the points A and B. Since this was asked about 20 days ago I'll go ahead and finish it up since it's fun.

\[A=1/2*b*h\] \[b=\sqrt{(x _{A}-x _{P})^2-(y _{A}-y _{P})^2}\] \[h=\sqrt{(x _{B}-x _{P})^2-(y _{B}-y _{P})^2}\]

Answer 3

Actually I lost interest, but it's pretty straight forward from here. Just plug in the values with y=x^2 and just take the derivative and set it equal to 0 to find the point where the area is largest between those two endpoints.