The following table represents the diameter of the cross section of a wire at continuous heights (feet) above the ground. Assume that each cross section is circular.

Height(ft) 2 6 10 14 18 22 26 30
Diameter 2 2 2.0 1.8 1.6 1.5 1.3 1.2 (feet)

(those should all line up)

a) Approximate how fast the diameter of the tree is changing 22 ft. above the ground. indicate units of measure

b) Use the trapezoid rule to approximate the volume of the tree from 14 ft to 30 ft above the ground. Indicate units of measure

C) the section of the tree from 2 ft to 8 ft is used to make a rectangular beam of length 6 ft. The strength of the beam varies jointly as its width and the square of its height. What should be the width and height of the beam in order to have the strongest beam?

a) To approximate how fast the diameter of the tree is changing 22 ft above the ground, we can calculate the difference in diameter between two consecutive heights and divide it by the difference in height.

Given the table, we can see that the diameter at 22 ft (1.5 ft) is smaller than the diameter at 26 ft (1.3 ft).

Difference in diameter = 1.5 ft - 1.3 ft = 0.2 ft
Difference in height = 26 ft - 22 ft = 4 ft

To find how fast the diameter is changing, we divide the difference in diameter by the difference in height:

Rate of change = (0.2 ft) / (4 ft) = 0.05 ft/ft

Therefore, the diameter of the tree is changing at a rate of 0.05 ft per foot of height.

b) To use the trapezoid rule to approximate the volume of the tree from 14 ft to 30 ft above the ground, we first need to calculate the average of the diameters at each pair of consecutive heights. Then, we can use the formula for the volume of a cylinder to find the volume of each small section and sum them up.

Let's calculate the average diameters:

Diameter at 14 ft = 1.8 ft
Diameter at 18 ft = 1.6 ft
Diameter at 22 ft = 1.5 ft
Diameter at 26 ft = 1.3 ft
Diameter at 30 ft = 1.2 ft

Using the trapezoid rule, we can approximate the volume of each small section as:

Volume of each section = (1/2) * (diameter1 + diameter2) * height

Now, we can calculate the volumes:

Volume of section from 14 ft to 18 ft = (1/2) * (1.8 ft + 1.6 ft) * 4 ft = 6.4 ft^3
Volume of section from 18 ft to 22 ft = (1/2) * (1.6 ft + 1.5 ft) * 4 ft = 6.2 ft^3
Volume of section from 22 ft to 26 ft = (1/2) * (1.5 ft + 1.3 ft) * 4 ft = 6.8 ft^3
Volume of section from 26 ft to 30 ft = (1/2) * (1.3 ft + 1.2 ft) * 4 ft = 5.0 ft^3

Finally, we can sum up the volumes of all sections:

Total volume from 14 ft to 30 ft = 6.4 ft^3 + 6.2 ft^3 + 6.8 ft^3 + 5.0 ft^3 = 24.4 ft^3

Therefore, the volume of the tree from 14 ft to 30 ft above the ground is approximately 24.4 cubic feet.

c) In order to have the strongest beam, we need to maximize the strength of the beam which varies jointly as its width and the square of its height.

Let's assume the width of the beam is "w" and the height is "h". And let the strength of the beam be represented by "S".

According to the given information, the strength of the beam varies jointly as its width and the square of its height. So, we can write the equation as:

S = k * w * h^2

where "k" is a constant of proportionality.

To find the dimensions (width and height) that maximize the strength of the beam, we need to optimize this equation.

Since the length of the beam is given as 6 ft, we can write another equation:

6 = w + 2h

To solve this system of equations, we can substitute the value of w from the second equation into the first equation:

S = k * (6 - 2h) * h^2

Now, we can maximize S by finding the critical points. To do this, we take the derivative of S with respect to h, set it equal to 0, and solve for h:

dS/dh = k * (6 - 2h) * 2h + k * h^2 * (-2) = 0

Expanding and simplifying the equation gives:

12h - 4h^2 - 2h^2 = 0

Simplifying further:

16h - 6h^2 = 0
2h(8 - 3h) = 0

Setting each factor equal to zero gives two possible solutions:

h = 0 or h = 8/3 ft

Since h = 0 is not a physically meaningful solution (beam with zero height), we discard it.

Therefore, the height of the beam that maximizes the strength is 8/3 ft.

To find the width, we substitute this value of h back into the second equation:

6 = w + 2 * (8/3)
6 = w + 16/3
w = 6 - 16/3
w = 6/3 - 16/3
w = -10/3
w = -3.33 ft (approximately)

However, width cannot be negative, so we discard this solution as well.

Therefore, there is no real-valued solution for the width.

In conclusion, to have the strongest beam, the height should be 8/3 ft and there is no real-valued solution for the width.