Math / Trigonometry

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1. A rectangular beam (a wooden pole) is to be cut from a cylindrical $\log$ of 20 in diameter as shown in the picture. (a) Show that the cross-sectional area of the rectangular beam (shown in the pictures below) is modeled by the function, $A(\theta)=200 \sin (2 \theta)$ (b) What is the maximum cross-sectional area of such a beam ?

Studdy Solution

STEP 1

What is this asking? We need to find a formula for the area of a rectangle cut out of a circular log and then figure out the biggest area we can make. Watch out! Remember that $\theta$ is only half of the rectangle's corner angle, and make sure your calculator is in degrees or radians as needed!

STEP 2

1. Define the function
2. Optimize the formula

STEP 3

Alright, let's dive into this!
We've got a $\textbf{20-inch diameter}$ log, and we're cutting a rectangle out of it.
The rectangle's width is $w$ and its height is $h$ .
We know the area of a rectangle is $A = w \cdot h$ .

STEP 4

Now, look at that angle $\theta$ !
It's tucked between the rectangle's diagonal and the circle's diameter.
Since the diameter of the log is $\textbf{20 inches}$ , the radius is $r = \frac{20}{2} = \textbf{10 inches}$ .

STEP 5

We can use some trigonometry magic!
Notice how $h$ and $w$ relate to the radius and $\theta$ .
We have $h = 2r \cdot \sin(\theta) = 2 \cdot 10 \cdot \sin(\theta) = 20 \cdot \sin(\theta)$ and $w = 2r \cdot \cos(\theta) = 2 \cdot 10 \cdot \cos(\theta) = 20 \cdot \cos(\theta)$ .

STEP 6

Let's plug those values into our area formula: $A = w \cdot h = (20 \cdot \cos(\theta)) \cdot (20 \cdot \sin(\theta)) = 400 \cdot \sin(\theta) \cdot \cos(\theta)$ .

STEP 7

Remember the double angle identity? $2 \cdot \sin(\theta) \cdot \cos(\theta) = \sin(2\theta)$ .
So, $A = 400 \cdot \sin(\theta) \cdot \cos(\theta) = 200 \cdot 2 \cdot \sin(\theta) \cdot \cos(\theta) = 200 \cdot \sin(2\theta)$ .
Boom! We just proved the area function is $A(\theta) = 200 \cdot \sin(2\theta)$ !

STEP 8

Now, how do we maximize the area?
Think about the sine function.
It reaches its maximum value of $\textbf{1}$ when the angle is $90^\circ$ or $\frac{\pi}{2}$ radians.

STEP 9

In our case, the angle inside the sine is $2\theta$ .
So, we need $2\theta = \frac{\pi}{2}$ , which means $\theta = \frac{\pi}{4}$ radians or $45^\circ$ .

STEP 10

Let's plug that back into our area formula: $A_{max} = 200 \cdot \sin(2 \cdot \frac{\pi}{4}) = 200 \cdot \sin(\frac{\pi}{2}) = 200 \cdot 1 = \textbf{200}$ .

STEP 11

The maximum cross-sectional area of the beam is $\textbf{200}$ square inches, which happens when $\theta$ is $45^\circ$ .

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Studdy Solution

STEP 1

What is this asking? We need to find a formula for the area of a rectangle cut out of a circular log and then figure out the biggest area we can make. Watch out! Remember that $\theta$ is only half of the rectangle's corner angle, and make sure your calculator is in degrees or radians as needed!

STEP 2

1. Define the function
2. Optimize the formula

STEP 3

Alright, let's dive into this!
We've got a $\textbf{20-inch diameter}$ log, and we're cutting a rectangle out of it.
The rectangle's width is $w$ and its height is $h$ .
We know the area of a rectangle is $A = w \cdot h$ .

STEP 4

Now, look at that angle $\theta$ !
It's tucked between the rectangle's diagonal and the circle's diameter.
Since the diameter of the log is $\textbf{20 inches}$ , the radius is $r = \frac{20}{2} = \textbf{10 inches}$ .

STEP 5

We can use some trigonometry magic!
Notice how $h$ and $w$ relate to the radius and $\theta$ .
We have $h = 2r \cdot \sin(\theta) = 2 \cdot 10 \cdot \sin(\theta) = 20 \cdot \sin(\theta)$ and $w = 2r \cdot \cos(\theta) = 2 \cdot 10 \cdot \cos(\theta) = 20 \cdot \cos(\theta)$ .

STEP 6

Let's plug those values into our area formula: $A = w \cdot h = (20 \cdot \cos(\theta)) \cdot (20 \cdot \sin(\theta)) = 400 \cdot \sin(\theta) \cdot \cos(\theta)$ .

STEP 7

STEP 8

Now, how do we maximize the area?
Think about the sine function.
It reaches its maximum value of $\textbf{1}$ when the angle is $90^\circ$ or $\frac{\pi}{2}$ radians.

STEP 9

In our case, the angle inside the sine is $2\theta$ .
So, we need $2\theta = \frac{\pi}{2}$ , which means $\theta = \frac{\pi}{4}$ radians or $45^\circ$ .

STEP 10

Let's plug that back into our area formula: $A_{max} = 200 \cdot \sin(2 \cdot \frac{\pi}{4}) = 200 \cdot \sin(\frac{\pi}{2}) = 200 \cdot 1 = \textbf{200}$ .

STEP 11

The maximum cross-sectional area of the beam is $\textbf{200}$ square inches, which happens when $\theta$ is $45^\circ$ .

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Studdy Solution

STEP 1

What is this asking? We need to find a formula for the area of a rectangle cut out of a circular log and then figure out the biggest area we can make. Watch out! Remember that θ\thetaθ is only half of the rectangle's corner angle, and make sure your calculator is in degrees or radians as needed!

STEP 2

1. Define the function2. Optimize the formula

STEP 3

Alright, let's dive into this!We've got a 20-inch diameter\textbf{20-inch diameter}20-inch diameter log, and we're cutting a rectangle out of it.The rectangle's width is www and its height is hhh.We know the area of a rectangle is A=w⋅hA = w \cdot hA=w⋅h.

STEP 4

Now, look at that angle θ\thetaθ!It's tucked between the rectangle's diagonal and the circle's diameter.Since the diameter of the log is 20 inches\textbf{20 inches}20 inches, the radius is r=202=10 inchesr = \frac{20}{2} = \textbf{10 inches}r=220​=10 inches.

STEP 5

STEP 6

Let's plug those values into our area formula: A=w⋅h=(20⋅cos⁡(θ))⋅(20⋅sin⁡(θ))=400⋅sin⁡(θ)⋅cos⁡(θ)A = w \cdot h = (20 \cdot \cos(\theta)) \cdot (20 \cdot \sin(\theta)) = 400 \cdot \sin(\theta) \cdot \cos(\theta)A=w⋅h=(20⋅cos(θ))⋅(20⋅sin(θ))=400⋅sin(θ)⋅cos(θ).

STEP 7

STEP 8

Now, how do we maximize the area?Think about the sine function.It reaches its **maximum value** of 1\textbf{1}1 when the angle is 90∘90^\circ90∘ or π2\frac{\pi}{2}2π​ radians.

STEP 9

In our case, the angle inside the sine is 2θ2\theta2θ.So, we need 2θ=π22\theta = \frac{\pi}{2}2θ=2π​, which means θ=π4\theta = \frac{\pi}{4}θ=4π​ radians or 45∘45^\circ45∘.

STEP 10

Let's plug that back into our area formula: Amax=200⋅sin⁡(2⋅π4)=200⋅sin⁡(π2)=200⋅1=200A_{max} = 200 \cdot \sin(2 \cdot \frac{\pi}{4}) = 200 \cdot \sin(\frac{\pi}{2}) = 200 \cdot 1 = \textbf{200}Amax​=200⋅sin(2⋅4π​)=200⋅sin(2π​)=200⋅1=200.

STEP 11

The maximum cross-sectional area of the beam is 200\textbf{200}200 square inches, which happens when θ\thetaθ is 45∘45^\circ45∘.

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

What is this asking? We need to find a formula for the area of a rectangle cut out of a circular log and then figure out the biggest area we can make. Watch out! Remember that $\theta$ is only half of the rectangle's corner angle, and make sure your calculator is in degrees or radians as needed!

1. Define the function
2. Optimize the formula

Alright, let's dive into this!
We've got a $\textbf{20-inch diameter}$ log, and we're cutting a rectangle out of it.
The rectangle's width is $w$ and its height is $h$ .
We know the area of a rectangle is $A = w \cdot h$ .

Now, look at that angle $\theta$ !
It's tucked between the rectangle's diagonal and the circle's diameter.
Since the diameter of the log is $\textbf{20 inches}$ , the radius is $r = \frac{20}{2} = \textbf{10 inches}$ .

Let's plug those values into our area formula: $A = w \cdot h = (20 \cdot \cos(\theta)) \cdot (20 \cdot \sin(\theta)) = 400 \cdot \sin(\theta) \cdot \cos(\theta)$ .

Now, how do we maximize the area?
Think about the sine function.
It reaches its maximum value of $\textbf{1}$ when the angle is $90^\circ$ or $\frac{\pi}{2}$ radians.

In our case, the angle inside the sine is $2\theta$ .
So, we need $2\theta = \frac{\pi}{2}$ , which means $\theta = \frac{\pi}{4}$ radians or $45^\circ$ .

Let's plug that back into our area formula: $A_{max} = 200 \cdot \sin(2 \cdot \frac{\pi}{4}) = 200 \cdot \sin(\frac{\pi}{2}) = 200 \cdot 1 = \textbf{200}$ .

The maximum cross-sectional area of the beam is $\textbf{200}$ square inches, which happens when $\theta$ is $45^\circ$ .