A rectangular beam is cut from a cylindrical log of radius 25 cm. The strength of a beam of width w and height h is proportional to wh2. (See Figure 4.70.) Find the width and height of the beam of maximum strength.

A rectangular beam is cut from a cylindrical log of radius r=10 cm. The strength S of a beam of width w and height h is given by

S=wh2
The cross section of the log and beam are shown in the figure below.

Determine the width and height of the beam of maximum strength that can be cut from the log. State your answers rounded to 2 decimal places.

Well, isn't this a tough nut to crack! Let's get things rolling.

So, we have this rectangular beam that's cut from a cylindrical log. The strength of the beam is proportional to wh^2, where w is the width and h is the height. We need to find the width and height of the beam that maximizes its strength.

To get started, let's assume the height of the beam is h, and the width is w. Now, we know that the beam is cut from a cylindrical log with a radius of 25 cm. So, the height of the beam (h) should be less than or equal to 2 times the radius (2 * 25 cm).

Since we want to maximize the strength, we need to maximize the product of wh^2. Now, we can express the product as P = wh^2.

To find the maximum value of P, we can take the derivative of P with respect to each variable (w and h) and set the derivatives equal to zero.

*derivative dance*

After some mathematical acrobatics, we'll find that the width (w) should be equal to the height (h).

So, the width and height of the beam of maximum strength are the same. In other words, we need a square-shaped beam!

But hey, don't let this square news bore you! Remember, it's all about maximizing that strength and keeping those logs in line. So, grab your saw and start cutting those beams. And don't forget to embrace the power of symmetry while you're at it!

I hope this solution brought a smile to your face, and may your beams be strong, sturdy, and full of charm!

To find the width and height of the beam of maximum strength, we need to maximize the function wh^2.

Let's first express the height of the beam in terms of the radius and width. Since the beam is cut from a cylindrical log, the height of the beam can be represented as twice the radius of the log, h = 2r.

Now we can rewrite the function in terms of w:
Strength = w * (2r)^2
= 4wr^2

To find the maximum strength, we need to find the critical points of the function. This can be done by taking the derivative of the strength function with respect to w and set it equal to zero:

d(Strength)/dw = 4r^2 = 0

From this, we can see that r = 0, which is not possible since the radius cannot be zero.

Thus, there are no critical points for the function. This means that the function does not have a maximum or minimum.

However, since the problem asks for the dimensions of the beam with the maximum strength, we can deduce that the maximum strength occurs when the width is as large as possible. Therefore, the width of the beam should be equal to the diameter of the log (twice the radius) for maximum strength.

Width (w) = 2r
= 2 * 25 cm
= 50 cm

Now that we have the width, we can substitute it back into the height equation:

Height (h) = 2r
= 2 * 25 cm
= 50 cm

Therefore, the width and height of the beam of maximum strength are both 50 cm.

To find the width and height of the beam that would result in maximum strength, we need to use the given information that the strength of the beam is proportional to wh^2.

Let's denote the width of the beam as w and the height as h. We want to maximize the strength, which means we need to find the values of w and h that maximize the function wh^2.

In this problem, we are told that the rectangular beam is cut from a cylindrical log of radius 25 cm. The rectangular beam will have the same width as the diameter of the log since the beam is cut from one end of the log to the other in a straight line. Therefore, the width of the beam is 2 * 25 cm = 50 cm.

Now, we need to determine the height of the beam. Since the beam is cut from a cylindrical log, the height of the beam can vary, but it cannot exceed the radius of the log. Therefore, the height of the beam, h, must be less than or equal to 25 cm.

To find the values of w and h that maximize the strength, we need to consider the function wh^2. Since we have a constraint on the height, h, we can rewrite the equation in terms of a single variable, w:

Strength = wh^2
Strength = w(25^2)
Strength = 625w

Now, we can see that the strength is directly proportional to the width of the beam, w. This means that as the width increases, the strength also increases. Since we want to maximize the strength, we want to choose the maximum possible value for w.

Therefore, the width of the beam of maximum strength is 50 cm, and the height can be any value between 0 cm and 25 cm.

Draw a diagram. Given an inscribed rectangle of width w and height h, the diagonal of the rectangle is a diameter of the circle.

w^2 + h^2 = d^2
h^2 = 2500 - w^2

s = kwh^2 = kw(2500-w^2) = 2500kw - kw^3
s' = 2500k - 3kw^2

max strength is where s' = 0

3w^2 = 2500
w = 28.87

Now you can find h.