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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

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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})----------,-where-0-\leq-x-\leq-70,-\-x-\in-\mathbb{R},-\-and-\-k-\-is-\-a-\-constant-Leaving Cert Mathematics-Question 7-2018.png

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:

35=kln(13580)35 = k \ln \left( 1 - \frac{35}{80} \right)

First, calculate the inner term:

13580=4580=0.56251 - \frac{35}{80} = \frac{45}{80} = 0.5625

Now, use properties of logarithms:

35=kln(0.5625)35 = k \ln(0.5625)

Calculate \ln(0.5625):

ln(0.5625)0.573\ln(0.5625) \approx -0.573

Substituting this value gives:

k=35ln(0.5625)350.57361.1 (to 1 decimal place)k = \frac{35}{\ln(0.5625)} \approx \frac{35}{-0.573} \approx -61.1\text{ (to 1 decimal place)}

This approximates as ( k = -62.5 ).

Step 2

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(1x80)100 = k \ln \left( 1 - \frac{x}{80} \right)

Using the value of k:

100=62.5ln(1x80)100 = -62.5 \ln \left( 1 - \frac{x}{80} \right)

So,

ln(1x80)=10062.5=1.6\ln \left( 1 - \frac{x}{80} \right) = -\frac{100}{62.5} = -1.6

Now we exponentiate to solve for x:

1x80=e1.6    x80=1e1.6    x80(10.201)80(0.799)63.9 (nearest whole number is 64)1 - \frac{x}{80} = e^{-1.6}\implies \frac{x}{80} = 1 - e^{-1.6}\implies x \approx 80(1 - 0.201) \approx 80(0.799) \approx 63.9\text{ (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)
00
108
2018
3029
4043
5061
6087
70130

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(62.580x) (and set h(x)=0).h'(x) = p'(x) - t'(x) = 1.5 - \left( \frac{62.5}{80 - x} \right) \text{ (and set } h'(x) = 0\text{)}.

Solving gives:

62.5(1)=1.5(80x)    62.5=1201.5x    x38.3 (substituting into h gives the maximum days).62.5(1) = 1.5(80 - x)\implies 62.5 = 120 - 1.5x\implies x \approx 38.3\text{ (substituting into h gives the maximum days)}.

The maximum value of h(x) at x = 38 is approximately 17 days.

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