To determine the cost of hiring the tractor for a distance of 80 kilometers, we need to derive the cost function based on the information provided. The cost function is composed of a constant part and a part that varies with the square of the distance covered. Let's denote the cost function as:
[tex]\[ C = a + b \cdot d^2 \][/tex]
where:
- [tex]\( C \)[/tex] is the cost of hiring the tractor,
- [tex]\( d \)[/tex] is the distance covered,
- [tex]\( a \)[/tex] is the constant part of the cost,
- [tex]\( b \)[/tex] is the coefficient for the distance squared part.
We have two data points:
1. When [tex]\( d = 10 \)[/tex] km, [tex]\( C = 3200 \)[/tex] ksh.
2. When [tex]\( d = 30 \)[/tex] km, [tex]\( C = 3450 \)[/tex] ksh.
Using these data points, we can set up two equations:
1. [tex]\( 3200 = a + b \cdot 10^2 \)[/tex]
2. [tex]\( 3450 = a + b \cdot 30^2 \)[/tex]
Let's solve these equations step by step.
First, rewrite the equations:
[tex]\[ 3200 = a + 100b \][/tex]
[tex]\[ 3450 = a + 900b \][/tex]
To eliminate [tex]\( a \)[/tex], subtract the first equation from the second:
[tex]\[ 3450 - 3200 = (a + 900b) - (a + 100b) \][/tex]
[tex]\[ 250 = 800b \][/tex]
Solving for [tex]\( b \)[/tex]:
[tex]\[ b = \frac{250}{800} \][/tex]
[tex]\[ b = 0.3125 \][/tex]
Now, substitute [tex]\( b \)[/tex] back into the first equation to solve for [tex]\( a \)[/tex]:
[tex]\[ 3200 = a + 100 \cdot 0.3125 \][/tex]
[tex]\[ 3200 = a + 31.25 \][/tex]
[tex]\[ a = 3200 - 31.25 \][/tex]
[tex]\[ a = 3168.75 \][/tex]
So, the cost function is:
[tex]\[ C = 3168.75 + 0.3125 \cdot d^2 \][/tex]
Finally, to determine the cost of hiring the tractor for 80 kilometers, substitute [tex]\( d = 80 \)[/tex] into the cost function:
[tex]\[ C = 3168.75 + 0.3125 \cdot 80^2 \][/tex]
[tex]\[ C = 3168.75 + 0.3125 \cdot 6400 \][/tex]
[tex]\[ C = 3168.75 + 2000 \][/tex]
[tex]\[ C = 5168.75 \][/tex]
Therefore, the cost of hiring the tractor for 80 kilometers is ksh 5,168.75.
