Question

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.

Answer 1

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 ?

Answer:

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