On the relation y=√(1-x^2) P is a point on the graph of this relation, A is the point (–1, 0) and B is the point (x, 0).

Question

Find an expression for the length PB in terms of x only and find the maximum area of the triangle ABP

Answer 1

P is a point on the top half of a circle with center at the origin

It helps to sketch the semi-circle and plot the points

y = sqr(1-x^2)

square both sides

y^2 =1 -x^2

x^2 + y^2 = 1^2, radius = 1, center = (0,0)

B is (1,0) the right end of the semi-circle

P is (0,1), is at the top of the semi-circle

A is (-1,0) is the left end of the semi-circle

APB is police terminology for All Points Bulliten (an irrelevant detail, although the circle suggests the possible limits of police radar to catch a speeding driver who is at a point P)

Triangle ABP is a right triangle with base = 2, height =1 and Area = bh/2 = 2(1)/2 = 1 = maximum area of a triangle inscribed inside the semi-circle, with top point P, base =AB, height =y, the y coordinate of point P

Max area = 1

any inscribed triangle has base =2, point P = (x,y) = (x, sqr(1-x^2), height = y = sqr(1-x^2)

and area = A = bh/2 = 2(sqr(1-x^2)/2 = sqr(1-x^2)

A = sqr(1-x^2)

to maximize A, take the derivative and set =0

A' = 2/sqr(1-x^2)(-2x) = 0

-4x/sqr(1-x^2) = 0

-4x = 0

x=0

y =sqr(1-x^2) = sqr(1-0) = sqr1 = 1

y=1, only at the top of the semi-circle is y=1

to maximize the triangle's area P = (0,1)