The time, in days of practice, it takes Jack to learn to type x words per minute (wpm) can be modelled by the function: t(x) = k ln tyn(1 - \frac{x}{80}) where 0 ≤ x ≤ 70, x ∈ R, and k is a constant - Leaving Cert Mathematics - Question 7 - 2018

To find the corresponding wpm for 100 days, we set up the equation:

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

Using k = -62.5,

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

Rearranging gives:

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

Exponentiating both sides results in:

$1 - \frac{x}{80} = e^{-1.6}$

Calculating gives:

$1 - \frac{x}{80} \approx 0.2019$

This implies:

$\frac{x}{80} \approx 0.7981$

So, $x \approx 80 \times 0.7981 \approx 63.85$

The nearest whole number is:

$x = 64 \, \text{wpm}$

To complete the table, we calculate t(x) for each value of x using:

t(x) = -62.5 \ln\left(1 - \frac{x}{80}\right).

Calculating values:

• For x = 0: t(0) = 0
• For x = 10: t(10) = 8
• For x = 20: t(20) = 18
• For x = 30: t(30) = 29
• For x = 40: t(40) = 43
• For x = 50: t(50) = 61
• For x = 60: t(60) = 87
• For x = 70: t(70) = 130

x (wpm)	t(x) (days)
0	0
10	8
20	18
30	29
40	43
50	61
60	87
70	130

Graph t(x) using these points.

The function p(x) = 1.5x can be graphed by plotting points:

• For x = 0, p(0) = 0
• For x = 10, p(10) = 15
• For x = 20, p(20) = 30
• For x = 30, p(30) = 45
• For x = 40, p(40) = 60
• For x = 50, p(50) = 75
• For x = 60, p(60) = 90
• For x = 70, p(70) = 105

Plot these points and connect them to form a straight line.

To find the solution to h(x) = 0, we look for the point where the graphs of p(x) and t(x) intersect. From the graph plotted, it appears that this intersection occurs at approximately:

$x \approx 62 \, \text{wpm}$

To find the maximum value of h(x), we need to differentiate h(x):

$h(x) = 1.5x - t(x)$

Now differentiate:

$h'(x) = 1.5 - t'(x)$

Using the chain rule and substituting t'(x), we set h'(x) to 0 and find critical points:

$h'(x) = 1.5 - \left(-62.5 \cdot \frac{-1}{80} \cdot \frac{1}{1 - \frac{x}{80}}\right)\cdots$

After simplification, solve for x to find the x-value that gives the maximum h(x), and then substitute back to find the maximum value.