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12345 [234]
4 years ago
13

g is constructing a triangular sail for a sailboat. Aluminum supports will run along the horizontal base and vertical height of

the sail, respectively. After making some measurements, Calculo determines that the height of the sail cannot exceed 12 feet, or else it will not be able to fit under the bridges in the area. In addition, the total weight of the supports cannot exceed 80 pounds, or else the boat will sink. If aluminum weighs 5 pounds per foot, what is the area of the largest sail Calculo can build? Fully justify your answer, doing each step you learned in the class prep. In particular, you must show that the area you found is indeed the largest possible.
Mathematics
1 answer:
valkas [14]4 years ago
7 0

Answer: A (max) = 1/2 ( 8 * 8 )   = 32 ft²

Step-by-step explanation:

We have two constrains

a) The height of the sail ( support) cannot exceed 12 feet

b) The total weight of the supports cannot exceed 80 pounds

The weight of a foot of aluminum is 5 .

If we call:

x lenght of base support,  y  height of the vertical support  and the sail shape (is a triangle) we have

A = 1/2* ( x + y )

We know that (given a constant perimeter rectangle, the maximun area  is the square

A = x * y                P = 2x +2y         y = (P - 2x ) ÷ 2  

A = x * ( P - 2x ) ÷ 2  ⇒    A = (Px -2x²) ÷ 2

if we get derivative      A´(x) = (P-4x)/2     ⇒  (P-4x)/2   = 0

P = 4x and    x = P/4

Now we can look an square as two straight triangles joined by the diagonal and these two triangles are of area maximun.

Therefore in our case the sail must be of 8 x 8 feet (base and height)

A (max) = 1/2 ( 8 * 8 )   = 32 ft²   and we will keep the condition of weigh

8 + 8 = 16 feet   and the supports weights  16 * 5  = 80 pounds

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3 years ago
I need help with letters (D) and (E). My model equation from letter (C) is: P = -55/4 t+ 340.
IceJOKER [234]

Answer:

(a) The two ordered pairs are (0 , 340) and (4 , 285)

(b) The slope is m = -55/4

The slope means the rate of decreases of the owl population was 55/4

per year (P decreased by 55/4 each year)

(c) The model equation is P = -55/4 t + 340

(d) The owl population in 2022 will be 216

(e)  At year 2038 will be no more owl in the park

Step-by-step explanation:

* Lets explain how to solve the problem

- The owl population in 2013 was measured to be 340

- In 2017 the owl population was measured again to be 285

- The owl population is P and the time is t where t measure the numbers

 of years since 2013

(a) Let t represented by the x-coordinates of the order pairs and P

   represented by the y-coordinates of the order pairs

∵ t is measured since 2013

∴ At 2013 ⇒ t = 0

∵ The population P in 2013 was 340

∴ The first order pair is (0 , 340)

∵ The time from 2013 to 2013 = 2017 - 2013 = 4 years

∴ At 2017 ⇒ t = 4

∵ The population at 2017 is 285

∴ The second order pair is (4 , 285)

* The two ordered pairs are (0 , 340) and (4 , 285)

(b) The slope of any lines whose endpoints are (x1 , y1) and (x2 , y2)

     is m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}

∵ (x1 , y1) is (0 , 340) and (x2 , y2) is (4 , 285)

∴ x1 = 0 , x2 = 4 and y1 = 340 , y2 = 285

∴ m = \frac{285-340}{4-0}=\frac{-55}{4}

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∵ The slope is negative value

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(c) The linear equation form is y = mx + c, where m is the slope and c is

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∵ The population is P and represented by y

∵ The time is t and represented by t

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∵ m = -55/4

∵ The initial amount of the population is 340

∴ P = -55/4 t + 340

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∵ t = 2022 - 2013 = 9 years

∵ P = -55/4 t + 340

∴ P = -55/4 (9) + 340 = 216.25 ≅ 216

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∵ P = 0

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- Add 55/4 t to both sides

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- Multiply both sides by 4

∴ 55 t = 1360

- Divide both sides by 55

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∵ 2013 + 25 = 2038

* At year 2038 will be no more owl in the park

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