Write the integer that represents the situation.

A norman window is constructed by adjoining a semicircle to the top of an ordinary rectangular. Find the dimensions of a norman

Yanka [14]

Answer:

$W\approx 8.72$ and $L\approx 15.57$ .

Step-by-step explanation:

Please find the attachment.

We have been given that a norman window is constructed by adjoining a semicircle to the top of an ordinary rectangular. The total perimeter is 38 feet.

The perimeter of the window will be equal to three sides of rectangle plus half the perimeter of circle. We can represent our given information in an equation as:

$2L+W+\frac{1}{2}(2\pi r)=38$

We can see that diameter of semicircle is W. We know that diameter is twice the radius, so we will get:

$2L+W+\frac{1}{2}(2r\pi)=38$

$2L+W+\frac{\pi}{2}W=38$

Let us find area of window equation as:

$\text{Area}=W\cdot L+\frac{1}{2}(\pi r^2)$

$\text{Area}=W\cdot L+\frac{1}{2}(\pi (\frac{W}{2})^2)$

$\text{Area}=W\cdot L+\frac{\pi}{2}(\frac{W}{2})^2)$

$\text{Area}=W\cdot L+\frac{\pi}{2}(\frac{W^2}{4})$

$\text{Area}=W\cdot L+\frac{\pi}{8}W^2$

Now, we will solve for L is terms W from perimeter equation as:

$L=38-(W+\frac{\pi }{2}W)$

Substitute this value in area equation:

$A=W\cdot (38-W-\frac{\pi }{2}W)+\frac{\pi}{8}W^2$

Since we need the area of window to maximize, so we need to optimize area equation.

$A=W\cdot (38-W-\frac{\pi }{2}W)+\frac{\pi}{8}W^2$

$A=38W-W^2-\frac{\pi }{2}W^2+\frac{\pi}{8}W^2$

Let us find derivative of area equation as:

$A'=38-2W-\frac{2\pi }{2}W+\frac{2\pi}{8}W$

$A'=38-2W-\pi W+\frac{\pi}{4}W$

$A'=38-2W-\frac{4\pi W}{4}+\frac{\pi}{4}W$

$A'=38-2W-\frac{3\pi W}{4}$

To find maxima, we will equate first derivative equal to 0 as:

$38-2W-\frac{3\pi W}{4}=0$

$-2W-\frac{3\pi W}{4}=-38$

$\frac{-8W-3\pi W}{4}=-38$

$\frac{-8W-3\pi W}{4}*4=-38*4$

$-8W-3\pi W=-152$

$8W+3\pi W=152$

$W(8+3\pi)=152$

$W=\frac{152}{8+3\pi}$

$W=8.723210$

$W\approx 8.72$

Upon substituting $W=8.723210$ in equation $L=38-(W+\frac{\pi }{2}W)$ , we will get:

$L=38-(8.723210+\frac{\pi }{2}8.723210)$

$L=38-(8.723210+\frac{8.723210\pi }{2})$

$L=38-(8.723210+\frac{27.40477245}{2})$

$L=38-(8.723210+13.70238622)$

$L=38-(22.42559622)$

$L=15.57440378$

$L\approx 15.57$

Therefore, the dimensions of the window that will maximize the area would be $W\approx 8.72$ and $L\approx 15.57$ .

8 0

2 years ago

Substitute the value in for the variable. Then solve. Remember your order of operations!

inna [77]

Answer:

4

Step-by-step explanation:

36/9 =4

When you see an x put 9 ;)

6 0

1 year ago

Read 2 more answers

The circumference of the circle-shaped rug is 69.08 inches.

adelina 88 [10]

We can find the diameter by dividing the circumference by 3.14
d = 22

Thus, our radius is 11.
r = 11

Now, let's use the area formula ( $3.14 \cdot r^2$ ) to solve for the area.
a = $3.14 \cdot (11^2) = \boxed{379.94}$

8 0

2 years ago

Keylee writes the equation 6x+12=2(3x+6), Can you find the number of solutions this equation has without solving for x? Explain

Natasha2012 [34]

Answer:

All Real Numbers Are The Solutions

4 0

2 years ago

The function s(t) represents the position of an object at time t moving along a line. Suppose s(2)=150 and s(5)=237. Find the a

trapecia [35]

Answer:

29 is answer.

Step-by-step explanation:

Given that the function s(t) represents the position of an object at time t moving along a line. Suppose s(2)=150 and s(5)=237.

To find average velocity of the object over the interval of time [1,3]

We know that derivative of s is velocity and antiderivative of velocity is position vector .

Since moving along a line equation of s is

use two point formula

$\frac{s-150}{237-150} =\frac{t-2}{5-2} \\s=29t-58+150\\s=29t+92$ gives the position at time t.

Average velocity in interval (1,3)

= $\frac{1}{3-1} (s(3)-s(1))\\=\frac{1}{2} [87+58-29-58]\\=29$

5 0

2 years ago