Question

Strength of a Beam A wooden beam has a rectangular cross section of height h in. and width w in. (see the accompanying figure). The strength S of the beam is directly proportional to its width and the square of its height. What are the dimensions of the cross section of the strongest beam that can be cut from a round log of diameter d = 26 in.? (Hint: S = kh^2w, where k is a constant of proportionality. Round your answers to one decimal place.) w = [ ] in h = [ ] in

          Strength of a Beam A wooden beam has a rectangular cross section of height h in. and width w in. (see the accompanying figure). The strength S of the beam is directly proportional to its width and the square of its height. What are the dimensions of the cross section of the strongest beam that can be cut from a round log of diameter d = 26 in.? (Hint: S = kh^2w, where k is a constant of proportionality. Round your answers to one decimal place.)

w = [ ] in
h = [ ] in

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Strength of a Beam A wooden beam has a rectangular cross section of height h in. and width w in. (see the accompanying figure). The strength S of the beam is directly proportional to its width and the square of its height. What are the dimensions of the cross section of the strongest beam that can be cut from a round log of diameter d = 26 in.? (Hint: S = kh^2w, where k is a constant of proportionality. Round your answers to one decimal place.)

w = [ ] in
h = [ ] in

Added by April R.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Eduard Sanchez

12/22/2022

Step 1

In this case, the diameter is 26 inches, so both the width and height must be less than or equal to 26 inches. Second, we know that the strength of the beam is directly proportional to its width and the square of its height. This means that to maximize the Show more…

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Transcript

00:01 Hi there, so for this problem we have the situation that is drawn in here.

00:05 Let's label this the width and h for the height of this.

00:11 And then we are given this distance right here that corresponds to the diameter of this, which is 22.

00:18 Now, once we know this, the question is, what are the dimensions of the conception of the strongest beam that can be caught from a road block of a diameter that is given of 22.

00:32 So first of all we consider that the we use the formula of the cross -sectional area which is equal to a common proportional decay times h to the square times the width and then let me just in here what we know from the figure that you can see in here is that the width a square plus the height to the square is equal to 22 to the square.

01:06 So what we can do in here is to, for example, so for h square, so h square is equal to 22 and that to the square minus the width to the square.

01:18 So we can substitute that into the surface area, into the area and expression.

01:25 So that will be the constant proportionality.

01:27 This times 22 to the square minus the width to the square and this times the width.

01:36 Now we just start simplifying this expression in here so we will find that this area is then equal to 22 to the square times the concept of proportionality times the width and this minus the concept of proportionality times the width to the three.

01:59 Okay, to the three in here.

02:03 Now with that, we can now, since we want to minima, yes, to maximize this function, we derivated with respect to the width.

02:20 So that will be 22 to the square times the constant of proportionality minus three times the concept of proportionality times the width a square...

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