trapezoid of greatest area?

Question

find the dimensions of the trapezoid of greatest area with longer base on the x axis and its other two vertices above teh x axis and on the graph of 4y = 16 - x (2) (that is x squared)

Answer 1

I think that you have to assume that the left hand upper vertex is on the y-axis such that the right hand upper axis is on the curve, re-written as y = -x^2/4 + 4. Looking at this equation, if y=0, then x=4; and if x=0, then y=4. So the most distal point on the curve "above" the x-axis has to be x=4. Then the vertices will be at (0,y), (sqr[1-y/4],y), (4,0), (0,0).
Then the area, A, under the curve is
A=y*sqr[1-y/4]+
+ (4-sqr[1-y/4])(y/2)
Now we differentiate A and set it equal to zero to get a max.
dA/dy = some number I am too tired to differentiate = 0
Solve for y. Put this value for y into the vertices in terms of y and you are done.

Answer 2

This is a really cool problem. The basic idea is to create an equation for the area of the trapezoid with respect to one variable, x. Then to find the max area, you derive the equation, set it equal to zero, and solve for x.

You might wanna graph the y = 4 - 1/4*x^2 first. It's a parabola opening below intersecting the x-axis at -4 and 4. So now you know the base of the trapezoid is 8. The length of the top of the trapezoid is going to be 2x, try to visualize it to see why. The height of the trapezoid is going to be 4 - 1/4*x^2, the equation given to you.

The area of a trapezoid is:
(base1+base2) / 2 * height

Plug it all in:
(8 + 2x) / 2 * (4 - 1/4*x^2)

Simplify:
area = -1/4*x^3 - x^2 + 4x + 16

Derive and set equal to zero:
-3/4*x^2 - 2x + 4 = 0

Quadratic Equation to solve for x:
x = 4/3

Now the final dimensions:
base1 = 8
base2 = 2x = 8/3
height = 4 - 1/4*x^2 = 32/9

Answer 3

You desire to find the inscribed trapezoid with the largest area in the region bounded by the x-axis and the curve 4y = 16 - x². It has one of its bases on the x-axis.

First find the zeroes for the curve 4y = 16 - x².

4y = 16 - x²
4y = (4 - x)(4 + x)
x = -4, 4

So the base on the x-axis is of length (4 - (-4)) = 8

4y = 16 - x²
y = 4 - x²/4

The width of the upper base b = 2x.
The height is h = y = 4 - x²/4.

A = area of trapezoid

A = (1/2)(8 + b)h = (1/2)(8 + 2x)(4 - x²/4)
A = (4 + x)(4 - x²/4) = 16 - x² + 4x - x³/4
A = 16 + 4x - x² - x³/4

Take the derivative to find the critical values.

dA/dx = 4 - 2x - (3/4)x² = 0
16 - 8x - 3x² = 0
3x² + 8x - 16 = 0
(3x - 4)(x + 4) = 0
x = -4, 4/3

Take the second derivative to find the nature of the critical values.

d²A/dx² = -2 - (3/2)x
= -2 - (3/2)(-4) = -2 + 6 = 4 > 0 implies relative minimum
= -2 - (3/2)(4/3) = -2 - 2 = -4 < 0 implies relative maximum

The dimensions of the trapezoid are

Bases are
8
2x = 2(4/3) = 8/3

Height is
h = y = 4 - (4/3)²/4 = 4 - 4/9 = 32/9

Base 1 = 8
Base 2 = 8/3
Height = 32/9

Answer 4

Let x be the half of the top base, and y be the height.

area = f(x) = (x+4)y
f'(x) = y+xy' = 4-x^2/4-(x+4)(x/2) = 0
3x^2+8x-16 = 0
(3x-4)(x+4) = 0
Solve for x,
x = 4/3
y = 4-(4/3)^2/4 = 32/9

Therefore, the dimensions of the trapezoid are:
base1 = 8/3
base2 = 8
height = 32/9