Question

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1. the cost c of producing n items varies directly as n and partly as the inverse of n.to produce two items it costs ksh.135 and to produce three items it costs ksh.140. calculate

a) the constant of proportionality and hence write the equation connecting c and n. (5marks)
b) the cost of producing 10 items (2marks)
c) the number of items produced at a cost of ksh.756. (3marks)

Explanation:

Part (a)

Step1: Define the cost function

Since the cost \( c \) varies directly as \( n \) and partly as the inverse of \( n \), we can write the equation as \( c = kn + \frac{m}{n} \), where \( k \) and \( m \) are constants of proportionality.

Step2: Substitute \( n = 2 \), \( c = 135 \)

Substituting into the equation: \( 135 = 2k + \frac{m}{2} \). Multiply through by 2 to get \( 270 = 4k + m \) (Equation 1).

Step3: Substitute \( n = 3 \), \( c = 140 \)

Substituting into the equation: \( 140 = 3k + \frac{m}{3} \). Multiply through by 3 to get \( 420 = 9k + m \) (Equation 2).

Step4: Solve the system of equations

Subtract Equation 1 from Equation 2: \( (9k + m) - (4k + m) = 420 - 270 \). Simplify: \( 5k = 150 \), so \( k = 30 \).

Step5: Find \( m \)

Substitute \( k = 30 \) into Equation 1: \( 270 = 4(30) + m \). Calculate \( 4(30) = 120 \), so \( m = 270 - 120 = 150 \).

Step6: Write the cost equation

Substitute \( k = 30 \) and \( m = 150 \) into \( c = kn + \frac{m}{n} \): \( c = 30n + \frac{150}{n} \).

Step1: Use the cost equation for \( n = 10 \)

We have the equation \( c = 30n + \frac{150}{n} \). Substitute \( n = 10 \): \( c = 30(10) + \frac{150}{10} \).

Step2: Calculate the cost

Calculate \( 30(10) = 300 \) and \( \frac{150}{10} = 15 \). Then \( c = 300 + 15 = 315 \).

Step1: Set up the equation for \( c = 756 \)

We use the equation \( c = 30n + \frac{150}{n} \). Substitute \( c = 756 \): \( 756 = 30n + \frac{150}{n} \).

Step2: Multiply through by \( n \)

Multiply each term by \( n \) to get a quadratic equation: \( 756n = 30n^2 + 150 \). Rearrange: \( 30n^2 - 756n + 150 = 0 \). Divide through by 6: \( 5n^2 - 126n + 25 = 0 \).

Step3: Solve the quadratic equation

Use the quadratic formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) for \( ax^2 + bx + c = 0 \). Here, \( a = 5 \), \( b = -126 \), \( c = 25 \).

Calculate the discriminant: \( \Delta = (-126)^2 - 4(5)(25) = 15876 - 500 = 15376 \). \( \sqrt{15376} = 124 \).

So \( n = \frac{126 \pm 124}{10} \).

First solution: \( n = \frac{126 + 124}{10} = \frac{250}{10} = 25 \).

Second solution: \( n = \frac{126 - 124}{10} = \frac{2}{10} = 0.2 \) (not valid as \( n \) represents number of items, must be positive integer).

Answer:

The constants are \( k = 30 \) and \( m = 150 \), and the equation is \( c = 30n + \frac{150}{n} \).

Part (b)