Mathematics, 20.11.2020 07:50 cuhh

MY NOTES Two stoves are located 45 feet apart, one giving out]] 4 times as much heat as the other. If you stand at a point on the line between the stoves at a distance x from the less powerful stove, the temperature of the air is given by H(x)=100 + 60 x2 + 240 (45−x)2 . Assume you move back an forth between the two stoves, always at least 4 ft from either stove.

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Answer from: kenyaa63

Missing Part of the question:

a) Assume you wish to determine the maximum and minimum temperatures you would experience. the domain to study for the function H(x) would be?

b)The is one critical number for the function on the domain in part a and it is?

c) The maximum temperature you would experience is ?

d) The minimum temperature you would experience is ?

a. Domain : $4 \leq x \leq 41$

b. Critical point: x = 12.33

c. The maximum is 115.09

d. The minimum is 100.62

Step-by-step explanation:

Given

$H(x) = 100 + \frac{60}{x^2} + \frac{240}{(45 - x)^2}$

Solving (a): The domain of H(x)

From the question, we understand that the distance between the two stores is 45ft

Let the first stove be at point A (0 ft) and

the second stove be at point B (45ft)

Since you move back and forth within 4ft from either stoves, then

Your maximum distance at A is (0 + 4) ft = 4ft

Your minimum distance at B is (45 - 4) ft = 41ft

Hence, the domain is: $4 \leq x \leq 41$

Solving (b): Critical Value

First, we have to differentiate H(x) w.r.t x

$H(x) = 100 + \frac{60}{x^2} + \frac{240}{(45 - x)^2}$

Differentiate

$H'(x) = 0 -\frac{120}{x^3} + \frac{480}{(45- x)^3}$

$H'(x) = -\frac{120}{x^3} + \frac{480}{(45- x)^3}$

Equate H'(x) to 0

$0 = -\frac{120}{x^3} + \frac{480}{(45- x)^3}$

Rewrite as:

$\frac{120}{x^3} = \frac{480}{(45- x)^3}$

Cross Multiply

480 * x^3 = 120 * (45 -x)^3

480 * x^3 = 120 * (45 -x)*(45 -x) * (45 -x)

480 * x^3 = 120 * (2025 -90x +x^2) * (45 -x)

480 * x^3 = 120 * (91125 -4050x +45x^2 -2025x - 90x^2 - x^3)

480 * x^3 = 120 * (91125 -4050x -2025x+45x^2 - 90x^2 - x^3)

480 * x^3 = 120 * (91125 -6075x-45x^2 - x^3)

Divide both sides by 120

4x^3 =91125 -6075x-45x^2 - x^3

4x^3 +x^3-91125 +6075x+45x^2= 0

5x^3-91125 +6075x+45x^2= 0

5x^3+45x^2 +6075x-91125= 0

Divide through by 5

x^3+9x^2 +1215x-18225= 0

Solving for x, we have that

$x\approx \:12.33067$

x = 12.33

Hence, the critical point is 12.33

Solving (x): Maximum temperature

Here, we simply substitute the endpoints of the domain in $H(x) = 100 + \frac{60}{x^2} + \frac{240}{(45 - x)^2}$

Let x = 4

$H(4) = 100 + \frac{60}{4^2} + \frac{240}{(45 - 4)^2}$

$H(4) = 100 + \frac{60}{16} + \frac{240}{(41)^2}$

$H(4) = 100 + \frac{60}{16} + \frac{240}{1681}$

H(4) = 100 + 3.75+ 0.14277215942

H(4) = 103.892772159

H(4) = 103.89 --- approximated

Let x = 41

$H(41) = 100 + \frac{60}{41^2} + \frac{240}{(45 - 41)^2}$

$H(41) = 100 + \frac{60}{41^2} + \frac{240}{4^2}$

$H(41) = 100 + \frac{60}{1681} + \frac{240}{16}$

H(41) = 100 + 0.03569303985 + 15

H(41) = 115.03569304

H(41) = 115.04 ---- approximated

Compare both values: The maximum is 115.09

Solving (d): The minimum temperature

In (b), we have that:

x = 12.33 --- critical point

The minimum occurs at this point

Substitute 12.33 for x in $H(x) = 100 + \frac{60}{x^2} + \frac{240}{(45 - x)^2}$

$H(12.33) = 100 + \frac{60}{12.33^2} + \frac{240}{(45 - 12.33)^2}$

$H(12.33) = 100 + \frac{60}{12.33^2} + \frac{240}{32.67^2}$

$H(12.33) = 100 + \frac{60}{152.0289} + \frac{240}{1067.3289}$

H(12.33) = 100 + 0.39466180443 + 0.22486039682

H(12.33) = 100.619522201

H(12.33) = 100.62 --- approximated

Hence, the minimum is 100.62

Answer from: Quest

b. the average rate of change of f(x) is less than the average rate of change of g(x).

step-by-step explanation:

since the length of the interval is the same in each case, we only need to compare the change in y-value to see which is greater.

f(-2) = -3f(1) = 6 . . . estimated from f(x) = (x+2)^2-3, which matches points on the graph (-1,-2) and (0, 1)

so the change in y-value for function f is (6 )) = 9.

the table tells us

g(-2) = 1g(1) = 31

so the change in y-value for function g is (31 -1) = 30.

the change in y-value of f(x) on the interval is less than the change in y-value of g(x) on the same interval, so the average rate of change of f(x) is less than the average rate of change of g(x).

Answer from: Quest

the red lines denote the population of the bacteria after 35 hours. that would be 2,409.24. the y-intercept is the blue line which is 800 because that is the starting point. the equation used to make the graph is 800(1.032)^h, where h is the number of hours.

step-by-step explanation:

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MY NOTES Two stoves are located 45 feet apart, one giving out]] 4 times as much heat as the other. I...

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