another calculus question....?

Question

find the area of the largest rectangle with lower base on the x-axis and upper vertices on the parabola y=16-5x2.

Answer 1

Find the area of the largest rectangle with lower base on the x-axis and upper vertices on the parabola

y=16 - 5x²

Axis of symmetry is y - axis with vertex at (0, 16) and concave down

Cuts x - axis (y = 0) at x = ± 4/√5 = ± 4√5 /5

So base of rectangle has limits -4√5 /5 < x < 4√5 /5

So let the coordinates of the basal corners be (-x, 0) and (x , 0)

So the height will be 16 - 5x²

So area = base * height
ie A = 2|x|(16 - 5x²)
= 32|x| - 10|x|³

dA/dx = 32 - 30x²
= 0 for stationary points

So x² = 32/30
= 16/15
Thus x = 16/15)
= ±4√15/ 15

So |x| = 4√15/ 15

So when x² < 32/30, dA/dx > 0 and when x² > 32/30, dA/dx < 0.
ie there is a maximum at x² = 32/30

So Amax = 2*4√15/ 15(16 - 5*32/30)
= 8√15/ 15 (16 - 32/6)
= 8√15/ 15 * 32/3)
= 256√15/ 45 units² (exactly)
(≈ 22.033 units²)

Answer 2

First make a sketch and draw the rectangle and parabola in.

The distance from the origin to the bottom right corner of the rectangle is equal to the x-value at that bottom right corner. Due to symmetry, the entire width of the rectangle is two times that or 2x.

According to the problem, both top corners are on the parabola. Therefore we know the height of the rectangle. It is just the y-value of the parabola at those points on the parabola. We know that you can find the height "y" of any point on the parabola if we know the "x" value because they give us the equation of the parabola. y=16-5x^2. Therefore the height of the rectangle is equal to 16-5x^2 because the x value at those points is just "x".

The area of a rectangle is width times height. So all we have to do to find the area is multiply the width, 2x, by the height, 16-5x^2.
This gives us A=32x-10x^3

Now here comes the calculus part. Since the problem asked for the "area of the largest rectangle", what we want to do is to maximize the area.

We have a formula for the area in terms of the x value: A=32x-10x^3 that we calculated previously.

Now find the x values that will give you the maximum value for A. We'll do this by differentiating A with respect to x and setting that equal to zero.
A'=32-30x^2=0

Solve

30x^2=32
x^2=16/15

x=+/- sqrt(16/15)

Now we have the x value that will give us the maximum area. All we have to do is plug this into our equation for area to calculate the area.

A=32x-10x^3
A=32*sqrt(16/15)-10(sqrt(16/15))^3

Whatever that value multiplies out to is your answer.

Answer 3

Equation of x-axis, y = 0
A line parallel to x-axis, y = k
Points of intersections of the parabola and the line y = k:
k = 16 - 5x^2
x^2 = (16-k)/5
x1 = sqrt[(16-k)/5] and x2 = - sqrt[(16-k)/5] = -x1

The rectangle is formed by the following 4 points:
(x1, 0), (x1, k), (-x1, k) and (-x1, 0)

The area of the rectangle
A = (2x1)(k)
= 2(sqrt[(16-k)/5])(k)
= 2(k)[sqrt[(16-k)/5]
= (2/sqrt5)(k)[sqrt[(16-k)]

To maximize, differentiate A with respect to k and let it be = 0
dA/dk = (2/sqrt5){sqrt[(16-k) - k . 1/2 .sqrt[1/ (16-k)]} = 0
2(16-k) - k = 0
32 - 2k - k = 0
k = 32/3
plugging value of k in A
A(max) = (2/sqrt5)(32/3)[sqrt(16 - 32/3)]
= (2/sqrt5)(32/3)[sqrt(16/3)]
= 256/sqrt(15)
= 66.1

Answer 4

Pick the up-right vertex point (x, y).
A (area) = 2xy = 2x(16-5x^2)

Solving A' = 0 for x gives at x = 1.0328, A(max) = 22.033 units^2