Answer:
The cars should be rented at $37 per day
The maximum income is $ 6845
Explanation:
To solve this problem, a model like this is proposed:
1. If for $ 1 increase in rate 5 fewers cars are rented, then for X monetary units 5X fewe cars are rented.
2. This expressed in rate and cars terms is:
Rate= 30 + x
Cars= ( 220 - 5x)
3. And in the income formula is:
Income=Rate X Cars
But the income formula in function of x monetary units [F(x)] is:
[tex]F(x)=(220-5x)(30+x)[/tex]
[tex]F(x)=6600-150x+220x-5x^{2}[/tex]
[tex]F(x)=-5x^{2} +70x+6600[/tex]
4. F (x) being a quadratic function, in the form [tex]f(x)=ax^{2} +bx+c[/tex], with [tex]a<0[/tex] , the vertex [tex](h,k)[/tex]is determined because is the point of maximum values of the function.
The value of x of the vertex will give us in how many monetary units can be increased the rent.
5. We find the value of h with the following formula:
[tex]h=\frac{-b}{2a}[/tex]
[tex]h=\frac{-70}{2(-5)}[/tex]
[tex]h=7[/tex]
6. We replace the value of h in rate formula:
Rate= 30+h
Rate= 30 + 7
Rate= 37
And we respond the first question: The cars should be rented at $37 per day
7. We replace the value of h in F(x) or Income formula:
[tex]F(x)=-5x^{2} +70x+6600[/tex]
[tex]F(x)=-5(7)^{2} +70(7)+6600[/tex]
[tex]F(x)=-245 +490+6600[/tex]
[tex]F(x)= 6845[/tex]
or
Income= Rate X Cars
Income = 37 X (220 - 5(37))
Income= 37 X 185
Income= 6845
And we respond the second question: The maximum income is $ 6845