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the curve with equation y=x(6-x) crosses the x-axis at the origin O and at the point R. the rectangle OPQR is such that PQ is tangent to the curve at its maximum point.
1. show that the area of the region between the curve and the x-axis is two thirds of the area of the rectangle OPQR.
a straight line through the origin meets the curve at the point A with x-coordinate a.
2. find, in terms of a, the area of the shaded region between the curve and the line. hence find, correct to three significant figures, the value of a for which this area is half the area of the rectangle OPQR.
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Hi annie26;
For 1)
R = 6, maximum occurs at x = 3 and y = 9. So the Area o f OPRQ is 54
The area under the curve is:
For 2)
When you draw the line to the curve at Point A with coordinates (a, 6a-a^2).
The area under the straight line is:
The area under the curve is:
So the area above the straight line and below the curve is B - A.
So a^3 / 6 is the area over the line and under the curve in terms of a.
Set that equal to 27 and solve.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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