The time, in days of practice, it takes Jack to learn to type x words per minute (wpm) can be modeled by the function:
t(x) = k ln
(1 - \frac{x}{80})
, where 0 \leq x \leq 70, \ x \in \mathbb{R}, \ and \ k \ is \ a \ constant - Leaving Cert Mathematics - Question 7 - 2018
Question 7
The time, in days of practice, it takes Jack to learn to type x words per minute (wpm) can be modeled by the function:
t(x) = k ln
(1 - \frac{x}{80})
... show full transcript
Worked Solution & Example Answer:The time, in days of practice, it takes Jack to learn to type x words per minute (wpm) can be modeled by the function:
t(x) = k ln
(1 - \frac{x}{80})
, where 0 \leq x \leq 70, \ x \in \mathbb{R}, \ and \ k \ is \ a \ constant - Leaving Cert Mathematics - Question 7 - 2018
Step 1
Based on the function t(x), Jack can learn to type 35 wpm in 35-96 days. Write the function above in terms of k and hence show that k = -62.5.
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Answer
To find k, substitute x = 35 and t(x) = 35 into the function:
Find the number of wpm that Jack can learn to type with 100 days of practice. Give your answer correct to the nearest whole number.
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Answer
To find the number of wpm, we set t(x) = 100:
100=kln(1−80x)
Using the value of k:
100=−62.5ln(1−80x)
So,
ln(1−80x)=−62.5100=−1.6
Now we exponentiate to solve for x:
1−80x=e−1.6⟹80x=1−e−1.6⟹x≈80(1−0.201)≈80(0.799)≈63.9 (nearest whole number is 64)
Step 3
Complete the table below, correct to the nearest whole number and hence draw the graph of t(x) for 0 ≤ x ≤ 70, x ∈ R.
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Answer
The table is filled using the function t(x).
x (wpm)
t(x) (days)
0
0
10
8
20
18
30
29
40
43
50
61
60
87
70
130
The graph is drawn using these points.
Step 4
A simpler function that could also be used to model the number of days needed to attain x wpm is p(x) = 1.5x. Draw, on the diagram above, the graph of p(x) for 0 ≤ x ≤ 70, x ∈ R.
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Answer
The graph of p(x) = 1.5x can be plotted from (0,0) to (70, 105).
Step 5
Let h(x) = p(x) - t(x). (i) Use your graphs above to estimate the solution to h(x) = 0 for x > 0.
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Answer
From the graphs plotted, the intersection point indicates approximately 62 wpm, where h(x) = 0.
Step 6
Let h(x) = p(x) - t(x). (ii) Use calculus to find the maximum value of h(x) for 0 ≤ x ≤ 70, x ∈ R. Give your answer correct to the nearest whole number.
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Answer
To find the maximum value, we differentiate h(x):
h′(x)=p′(x)−t′(x)=1.5−(80−x62.5) (and set h′(x)=0).
Solving gives:
62.5(1)=1.5(80−x)⟹62.5=120−1.5x⟹x≈38.3 (substituting into h gives the maximum days).
The maximum value of h(x) at x = 38 is approximately 17 days.
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