## Related questions with answers

Citywide Ministries plans a five-day ski trip for children. The children may ski from one to five of the days, but the number scheduled to ski each day is the following:

$\begin{matrix} \text{Day} & \text{No. of Skiers}\\ \text{Monday} & \text{15}\\ \text{Tuesday} & \text{22}\\ \text{Wednesday} & \text{19}\\ \text{Thursday} & \text{24}\\ \text{Friday} & \text{31}\\ \end{matrix}$

Citywide Ministries can obtain two-day lift tickets for $36 each or three-day lift tickets for$47. Find the number of tickets they should purchase for each two-day period and each three-day period so that each child can ski on the day scheduled and that minimizes the cost of lift tickets.

Solution

VerifiedFirstly we are going to set the substitution. We are going to determine the possible ways to buy a lift ticket for two days or three days in the span of five days. Since there are $4$ ways to buy a two-day lift ticket and $3$ ways to buy a three-day lift ticket, we are going to label those ways as $x_1$, $x_2$, $x_3$, $x_4$, $x_5$, $x_6$, and $x_7$ where $x_1$, $x_2$, $x_3$, and $x_4$ represent the lift tickets bought for first and second, second and third, third and fourth, and fourth and fifth day respectively, and $x_5$, $x_6$, and $x_7$ represent the projectors bought in a first, second and third, a second, third, and fourth, and a third, fourth, and fifth day respectively.

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