extension question... help please!!! parabola -?

Question

parabola y=4-(x)^2 cuts x axis at R and S... P (x,y) lies on parabola in 1st quadrant, and Q lies on parabols such that PQ is paralel to the x-axis...

i already worked out R is (-2,0) and S is (2,0)

now i need to show area of trapeziumb PQRS is given by:
A = (2-x)(4-x^2)
aaargh!!??? wat!? Area of trapezium = 1/2(a + b)h... but not much help?

then i need to find value of x which gives the maximum value of A... but i think i can work that out if i get this.

pls help, thanx heaps :)

so is area max when x = 2/3? because ther was an error in the ssecond persons working out...
please confirm or show how 2 do :)
thx

Answer 1

You have the point R (-2, 0) and the point S (2, 0). This forms the base with a distance of 4.

Now you have a point P(x, y)... that makes the height y (because it is y units above the x-axis). y - 0 = y

Point Q will be at (-x, y) because the function is symmetric around the y-axis.

a = Distance of RS = 4
b = Distance of PQ = 2x
h = Height = y (which is the same as 4 - x², from your equation.)

Using the trapezium formula:
Area = ½(a + b)h
Area = ½(4 + 2x)(4 - x²)

Now you just distribute the ½ through the first set of parentheses and you have exactly what you need:

Area = (2 + x)(4 - x²)

Try the last bit on your own, but if you need more help... post additional notes on your question.

Answer 2

You typed the area WRONGLY!
It is (2 + x)(4 - x^2) not "A = (2-x)(4-x^2)".

The value of A is indeed maximum at x = 2/3 and
A_max = 256/27.

When you get x = 2/3, -2 from dA/dx = 0 then you can simply discard the negative because P has to lie in 1st quadrant so its abscissa cannot be negative.

Answer 3

Let P = (x, 4 - x^2)
Then Q = (-x, 4 - x^2)
Area of trapezium PQRS,
A = (1/2)(4 + 2x)(4 - x^2)
= (2 + x)(4 - x^2)
= - x^3 -2x^2 + 4x + 8.
[Area mentioned by you is incorrect.]

For maximum area, dA/dx = 0 and d2A/dx2 < 0
dA/dx = - 3x^2 - 4x + 4 = 0
=> 3x^2 + 4x - 4 = 0
=> x = (1/6) (- 4 ± √(16 + 48)) = 2/3 or - 2
d2A/dx2 = - 6x - 4 > 0 for x = 2 and < 0 for x = 2/3
=> Max. area = (2 + 2/3)(4 - (4/9)) = 256/27.