## Chapter 43Solutions for the fourteenth week’s assignments

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### 43.1 Exercises for 7.3

Problem 25
The ratio is {2.5\text{ oz oil}\over 1\text{ gal gas}}. So given 2.75 gallons, {2.5\text{ oz oil}\over 1\text{ gal gas}} ⋅ 2.75\text{ gal gas} = \mathbf{6.875} ounces of oil are required.
Problem 26
Here the ratio is {5.5\text{ oz oil}\over 1\text{ gal gas}}. Given 22 ounces of oil, {1\text{ gal gas}\over 5.5\text{ oz oil}} ⋅ 22\text{ oz oil} = \mathbf{4} gallons of gas are required.
Problem 68
“Varies directly” implies a proportional relationship. Here the ratio is {5\text{ psi}\over 200\text{ deg K}}. So at 300 degrees Kelvin, the pressure is {5\text{ psi}\over 200\text{ deg K}} ⋅ 300\text{ deg K} = \mathbf{7.5\text{ psi}}.
Problem 70
The correct ratio here is {12\text{ pounds}\over 3\text{ in}} , and the force to compress 5 inches is {12\text{ pounds}\over 3\text{ in}} ⋅ 5\text{ in} = \mathbf{20\text{ pounds}}.

### 43.2 Exercises for 7.4

Problem 64
The target heart rate for a 35 year old lies in [129.5,157.25]. Rounding this to nearest even (safer in general) gives the interval \mathbf{[130,157]}. When dealing with intervals like these, however, rounding each endpoint to nearest is not necessarily the best idea. In this case, the result was within the original interval, so all numbers in the rounded interval still satisfy the relationship. If we had rounded 129.5 to 129, then the lower endpoint would not have satisfied the relationship. Which direction to round depends on the problem and assumptions in the model1 . Your age: Likely is less than mine, which in turn is less than the first part. Note that you can treat each side, .7(220 - A) and .85(220 - A), as a line. Both have negative slopes, so both decrease with increasing age.
Problem 65
At break-even, the cost C is equal to the revenue R. So at break-even, 20x + 100 = 24x and x = 25. Now the question becomes on which side the company shows a profit, or R - C > 0. Substituting for R and C, (24x) - (20x + 100) > 0 or x > 25. The smallest whole number of units x to show a profit is then 26 and not 25.
Problem 66
Here R = 5.5x and C = 3x + 2300, so R - C > 0 becomes 5.5x - (3x + 2300) > 0 or 2.5x > 2300 and x > 920. So the smallest profitable x is 921.

### 43.3 Exercises for 7.5

When what I say clearly does not apply to the problem, you should tell me. Here I assigned Problem 60 in Section 7.5 when I meant Section 7.7. I’ve included the answer for the problem in Section 7.5, even though it’s a pointless problem.

Problem 60
{({p}^{-1})}^{3}{p}^{-4} = {p}^{-3}{p}^{-4} = \mathbf{{p}^{-7}}

### 43.4 Exercises for 7.7

Problem 60 by plotting
The function to plot is y = {x}^{2} + {(x + 30)}^{2} - 15{0}^{2} = 2{x}^{2} + 60x - 21600. Evaluating at the points x \{80,85,90,95,100\} gives:

 x 80 85 90 95 100 y -4000 -2050 0 2150 4400

So the zero is at x = 90. Plotting these segments:

roblem 60 by Pythagorean thm
By the Pythagorean theorem, {x}^{2} + {(x + 30)}^{2} = 15{0}^{2}. Then we need to find roots of 2{x}^{2} + 60x - 21600 = 2({x}^{2} + 30x - 10800). We can solve this by simply trying points or by the quadratic equation. For the latter, we can use the simpler {x}^{2} + 30x - 10800 and find
\eqalignno{ x & = {-30 ±\sqrt{3{0}^{2 } + 4 ⋅ 10800}\over 2} & & \cr & = {-30 ± 210\over 2} = -15 ± 105 & & \cr & = 90\text{ or } - 120. & & }

For using points, find convenient numbers in the problem and try them. Here, to numbers that pop out are 30 and 150. Evaluating at each gives (30,-14400) and (150,50400), so we know the zero must be somewhere between them. Half-way is (30 + 150)∕2 = 90, so trying 90 finds the solution.

### 43.5 Exercises for 8.1

Problem 56, using the point form of the line
The two points closest to 1985 are (1980,4.5) and (1990,5.2). The point form here is
 {x - 1980\over 1990 - 1980} = {y - 4.5\over 5.2 - 4.5}.

Plugging in x = 1985 and solving for y gives y = \mathbf{4.85} million. Repeating around 1995,

 {x - 1990\over 2000 - 1990} = {y - 5.2\over 5.8 - 5.2}.

Using x = 1995 gives y = \mathbf{5.5} million.

### 43.6 Exercises for 8.3

Problem 70
(This was the “bad graph” example I used previously to show how a linear relationship can be hidden. We can start with the point form,
 {x - 150\over 1400 - 150} = {y - 5000\over 24000 - 5000}.

Solving for y gives the slope-intercept form,

 y = 15.2x + 2720.

To verify this, check that x = 150 gives y = 5000, and x = 1400 gives y = 24000.

Problem 72
Part a
Again, starting from point form is easiest:
 {x - 5\over 7 - 5} = {y - 24075\over 26628 - 24075}.

Solving for y,

\eqalignno{ y & = 24075 + {2553\over 2} (x - 5) & & \cr & = 1276.5x + 17692.5. & & }
Part b
The slope from 1995 to 1997 is positive, but the graph shows a negative slope from 1993 or 1994 to 1995. Thus the linear model above will not approximate those well. The graph shows non-linear variation, so there is no reason to expect the 1995–1997 line to continue to 1998, so no to all.
Problem 74
Part a and b
The points in question are \mathbf{(0,32)} and \mathbf{(100,212)}.
Part c
The slope is {Δy\over Δx} = {21{2}^{∘}\text{F}-3{2}^{∘}\text{F}\over 10{0}^{∘}\text{C}-{0}^{∘}\text{C}} = \mathbf{{{9}^{∘}\text{F}\over { 5}^{∘}\text{C}}}.
Part d
The point (0,32) provides the y-intercept, so the line is \mathbf{y = {{9}^{∘}\text{F}\over { 5}^{∘}\text{C}}x + 3{2}^{∘}\text{F}}.
Part e
A function of x in terms of y: \mathbf{x = {{5}^{∘}\text{C}\over {9}^{∘}\text{F}}(y - 3{2}^{∘}\text{F})}.
Part f
The graph shows that 5{0}^{∘}\text{C} is 12{2}^{∘}\text{F}. You can tell that the graph is of the conversion {}^{∘}\text{C} {→}^{∘}\text{F} because the y-intercept is positive.

### 43.7 Exercises for 8.6

Problem 50
At 7% compounded quarterly2 , $60 000 will grow to in$60\kern 1.66702pt 000 ⋅ {(1 + {0.07\over 4} )}^{20} ≈ $84886.69 5 years. Compounded “continuously”3 , the same amount will grow to$60\kern 1.66702pt 000 ⋅ {e}^{0.0675⋅5} ≈ $84086.38. Here the higher rate compounded quarterly is the better one, with a difference of approximately$800.32.

### 43.8 Exercises for 8.7

Problem 50
The nonsensical system here is W = L - 44\text{ ft} and 2L + 2W = 288\text{ ft}. Solving by substituting the former into the latter gives L = 144\text{ ft} - W = 144\text{ ft} - L + 44\text{ ft}, so L = \mathbf{94\text{ ft}}. Now W = L - 44\text{ ft} = \mathbf{50}\text{ ft}.
Problem 78
The first statement gives {150\text{ km}\over T\text{ km/hr}} = {400\text{ km}\over P\text{ km/hr}} or equivalently 150\text{ km} ⋅ P\text{ km/hr} = 400\text{ km} ⋅ T\text{ km/hr}. The second statement gives P\text{ km/hr} = 3T\text{ km/hr} - 20\text{ km/hr}. Solving, the plane’s speed is T\text{ km/hr} = \mathbf{60}\text{ km/hr}, and the train’s speed is P\text{ km/hr} = \mathbf{160}\text{ km/hr}. Plugging these into either of the given equations verifies the solution.
Problem 86
Translating into algebra,
\begin{array}{cl} C = A + B + 10, & \\ B = 2A,\text{ and} & \\ A + B + C = 490.& \end{array}

This is best solved by substitution. Written in terms of C, the last equation becomes (C - 10) + C = 490, or \mathbf{C = 250}. Now writing the first in terms of A after substituting C, 250 = A + 2A + 10 or \mathbf{A = 80}. Now \mathbf{B = 160}. Substituting these into the equations above verifies this solution.

### 43.9 Exercises for 8.8

Problem 30
Let A and B be the servings of each product. Summarizing the information:

 Product I (g/serving) II (g/serving) cost (/serving) A 3 2 0.25 B 2 4 0.40 So the function to minimize is the cost 0.25A + 0.40B. The constraints are that there be at least 15 g of I, or 3A + 2B ≥ 15, and 15 g of II, or 2A + 4B ≥ 15. There also are trivial constraints A ≥ 0 and B ≥ 0. The problem to solve is to \eqalignno{ {\mathop{min}}_{A,B}\quad &0.25A + 0.40B & & \cr \text{subject to}\quad &3A + 2B ≥ 15, & & \cr &2A + 4B ≥ 15, & & \cr &A ≥ 0,\text{ and} & & \cr &B ≥ 0. & & \cr & & } The first two lines intersect at (3.75,1.875). The A intercepts are 5 and 7.5, and only (0,7.5) is feasible. The B intercepts are 7.5 and 3.75, and only (7.5,0) is feasible. So the points to check and the function values are  Point Cost (3.75,1.875)1.6875 (0,7.5) $3 (7.5,0)$1.875

So the cheapest combination is 3.75 of A and 1.875 of B at $1.6875. Problem 34 Summarizing and assigning variables:  Kind Variable Oven time (hr) Decorating time (hr) Profit ($) Cookie C 1.5 {2\over 3} 20 Cake A 2 3 30

So the problem is to

\eqalignno{ {\mathop{max}}_{C,A}\quad &20C + 30A & & \cr \text{subject to}\quad &1.5C + 2A ≤ 15, & & \cr &{2\over 3}C + 3A ≤ 13, & & \cr &C ≥ 0,\text{ and} & & \cr &A ≥ 0. & & }

The interesting points and their values:

 Point Profit ($) (0,0) 0 (6,3) 210 (0, {13\over 3} ) 130 (10,0) 200 So the best combination is 6 cookie batches and 3 cake batches for a profit of$210.