The problem describes a company that has developed a new product and wants to determine the optimal selling price. The provided table shows the relationship between the selling price $x_i$ (in thousands of CFA francs) and the number of units customers are willing to buy $y_i$. The design costs are 28 million CFA francs, and the production cost per unit is 25,000 CFA francs. The daily production $h(v)$ depends on the machine's speed $v$ (in km/h), and the function is given by $h(v) = -v + 3000 - \frac{8100}{v}$, where $v$ is between 50 and 120. There are two tasks: 1. Determine the speed $v$ that maximizes daily production and calculate the number of units produced at that speed.

1. Problem Description

The problem describes a company that has developed a new product and wants to determine the optimal selling price. The provided table shows the relationship between the selling price $x_i$ (in thousands of CFA francs) and the number of units customers are willing to buy $y_i$. The design costs are 28 million CFA francs, and the production cost per unit is 25,000 CFA francs. The daily production $h(v)$ depends on the machine's speed $v$ (in km/h), and the function is given by $h(v) = -v + 3000 - \frac{8100}{v}$, where $v$ is between 50 and

1

2

0. There are two tasks:

1. Determine the speed $v$ that maximizes daily production and calculate the number of units produced at that speed.

2. Use linear regression (least squares method) to determine the selling price $x$ that maximizes profit.

2. Solution Steps

Task 1: Maximizing Production

We want to maximize $h(v) = -v + 3000 - \frac{8100}{v}$ for $v \in [50, 120]$.

To find the maximum, we can take the derivative of $h(v)$ with respect to $v$ and set it equal to

0.

$h'(v) = -1 + \frac{8100}{v^2}$

Setting $h'(v) = 0$:

$-1 + \frac{8100}{v^2} = 0$

$\frac{8100}{v^2} = 1$

$v^2 = 8100$

$v = \pm 90$

Since $v$ must be positive, $v = 90$.

Since $v=90$ is within the interval $[50, 120]$, it is a valid critical point.

To confirm that this is a maximum, we can check the second derivative:

$h''(v) = -\frac{16200}{v^3}$

Since $v > 0$, $h''(v) < 0$, which means that $v = 90$ corresponds to a maximum.

Now we calculate the daily production at $v = 90$:

$h(90) = -90 + 3000 - \frac{8100}{90} = -90 + 3000 - 90 = 2820$

Task 2: Linear Regression and Profit Maximization

First, we will determine the linear regression equation.

The data points are $(x_i, y_i)$, where $x_i$ is the selling price and $y_i$ is the number of units sold. We want to find a line of the form $y = ax + b$ that best fits the data.

We need to calculate the following sums:

$\sum x_i, \sum y_i, \sum x_i y_i, \sum x_i^2$, and $n$ (number of data points).

$x_i = [60, 80, 100, 120, 140, 160, 180, 200]$

$y_i = [952, 805, 630, 522, 510, 324, 205, 84]$

$n = 8$

$\sum x_i = 60 + 80 + 100 + 120 + 140 + 160 + 180 + 200 = 1040$

$\sum y_i = 952 + 805 + 630 + 522 + 510 + 324 + 205 + 84 = 4032$

$\sum x_i y_i = (60*952) + (80*805) + (100*630) + (120*522) + (140*510) + (160*324) + (180*205) + (200*84) = 57120 + 64400 + 63000 + 62640 + 71400 + 51840 + 36900 + 16800 = 424100$

$\sum x_i^2 = 60^2 + 80^2 + 100^2 + 120^2 + 140^2 + 160^2 + 180^2 + 200^2 = 3600 + 6400 + 10000 + 14400 + 19600 + 25600 + 32400 + 40000 = 152000$

Now we can calculate $a$ and $b$:

$a = \frac{n(\sum x_i y_i) - (\sum x_i)(\sum y_i)}{n(\sum x_i^2) - (\sum x_i)^2} = \frac{8(424100) - (1040)(4032)}{8(152000) - (1040)^2} = \frac{3392800 - 4193280}{1216000 - 1081600} = \frac{-800480}{134400} \approx -5.956$

$b = \frac{(\sum y_i) - a(\sum x_i)}{n} = \frac{4032 - (-5.956)(1040)}{8} = \frac{4032 + 6194.24}{8} = \frac{10226.24}{8} \approx 1278.28$

So, the linear regression equation is $y = -5.956x + 1278.28$

Now we need to find the profit function. The revenue is $x * y = x(-5.956x + 1278.28) = -5.956x^2 + 1278.28x$ (in thousands of CFA francs).

The total cost is $28,000,000 + 25,000y = 28,000,000 + 25,000(-5.956x + 1278.28)$ (in CFA francs).

Converting costs to thousands of CFA francs and combining terms: $28000 + 25(-5.956x + 1278.28) = 28000 - 148.9x + 31957 = 59957-148.9x$.

The profit $P(x)$ is revenue minus cost:

$P(x) = (-5.956x^2 + 1278.28x) - (28000 + 25y) = -5.956x^2 + 1278.28x -(59957 - 148.9x)= -5.956x^2 + 1427.18x - 59957$

To maximize the profit, we take the derivative and set it equal to 0:

$P'(x) = -11.912x + 1427.18 = 0$

$11.912x = 1427.18$

$x = \frac{1427.18}{11.912} \approx 119.80$

3. Final Answer

1. The machine should operate at a speed of $v = 90$ km/h for maximum daily production. The number of units produced at this speed is $h(90) = 2820$.

2. The selling price that maximizes profit is approximately $x = 119.80$ thousand CFA francs (or 119,800 CFA francs).