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Answer:
Number of student tickets were sold = 409
Step-by-step explanation:
Total number of tickets sold = 771
I.e Students ticket (S) + Non students ticket(N S) = 771
Total amount to be paid for tickets = $ 3037
Amount paid by students per tickets = $3
Amount paid by non-students per tickets = $5
So, according to question
S + NS = 771
3 S + 5 NS = 3037
Solve both equations
5 S + 5 NS = 771 × 5
3 S + 5 NS = 3037
so ,
(5 S + 5 NS) - (3 S + 5 NS ) = 3855 - 3037
Or, 2 S = 818
I.e S = = 409
Hence, The number of student tickets were sold = 409 Answer
Given that for each <span>$2 increase in price, the demand is less and 4 fewer cars are rented.
Let x be the number of $2 increases in price, then the revenue from renting cars is given by
.
Also, given that f</span><span>or each car that is rented, there are routine maintenance costs of $5 per day, then the total cost of renting cars is given by
Profit is given by revenue - cost.
Thus, the profit from renting cars is given by
</span><span>
For maximum profit, the differentiation of the profit function equals zero.
i.e.
</span><span>
The price of renting a car is given by 48 + 2x = 48 + 2(8) = 48 + 16 = 64.
Therefore, the </span><span>rental charge will maximize profit is $64.</span>
Let x be the number of $2 increases in price, then the revenue from renting cars is given by
Also, given that f</span><span>or each car that is rented, there are routine maintenance costs of $5 per day, then the total cost of renting cars is given by
Profit is given by revenue - cost.
Thus, the profit from renting cars is given by
</span><span>
For maximum profit, the differentiation of the profit function equals zero.
i.e.
</span><span>
The price of renting a car is given by 48 + 2x = 48 + 2(8) = 48 + 16 = 64.
Therefore, the </span><span>rental charge will maximize profit is $64.</span>