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This DE has characteristic equation
with a repeated root at r = 3/2. Then the characteristic solution is
which has derivative
Use the given initial conditions to solve for the constants:
and so the particular solution to the IVP is
(A) We let y = the cost and are told x = the number of people. Since you pay $20 per person, the cost is 20x. That is, y=20x
(B) Again, let the cost =y and the number of people is given as x. You pay $10 per person or 10x plus an additional $50 for the room. That is, y=10x+50
(c) Link to graphs: https://www.desmos.com/calculator but if that doesn't work see the attachment for a screen shot. You just have to put the equations (type them) at left and the graph comes automatically.
(D) The admission price is the same when the two equations are equal. You can find this by setting them equal to each other as such: 20x = 10x+50 and solving for x. However, since you just graphed them the point of intersection (where the lines share/have the same point) gives the information. Remembers that (x,y) = (people, cost). The graphs intersect at (5, 100) so for 5 people the cost is the same and the cost is $100.
(E) For the regular rate we let x = 6 and solve for y (the cost). We get y = 20x which is y = (20)(6)=120. It costs $120 using the regular rate to take 6 people. Now let's use the equation for the group rate again with x = 6. Here we get y = 10x +50 or y = 10(6)+50 = $110. The group rate costs $110.
(F) The cost is the same at 5 people but if there are more than five the group rate is better as we saw in part E. So the regular rate is better for less than 5 people.
(G) Here y = $150. Let us use the group rate formula and solve for x (the number of people). 10x+50 = 150 so 10x = 100 and x = 10. Since 10 is more than 5 this is the better deal. However if you don't believe it or want to double check we can solve for x using y = 150 and the regular rate equation. We get: 20x = 150 so x = 7.5 Since we can't bring half a person we would only be able to bring 7 and that is less than 10 so this is not the best choice. Use the group rate and bring 10 people!
(B) Again, let the cost =y and the number of people is given as x. You pay $10 per person or 10x plus an additional $50 for the room. That is, y=10x+50
(c) Link to graphs: https://www.desmos.com/calculator but if that doesn't work see the attachment for a screen shot. You just have to put the equations (type them) at left and the graph comes automatically.
(D) The admission price is the same when the two equations are equal. You can find this by setting them equal to each other as such: 20x = 10x+50 and solving for x. However, since you just graphed them the point of intersection (where the lines share/have the same point) gives the information. Remembers that (x,y) = (people, cost). The graphs intersect at (5, 100) so for 5 people the cost is the same and the cost is $100.
(E) For the regular rate we let x = 6 and solve for y (the cost). We get y = 20x which is y = (20)(6)=120. It costs $120 using the regular rate to take 6 people. Now let's use the equation for the group rate again with x = 6. Here we get y = 10x +50 or y = 10(6)+50 = $110. The group rate costs $110.
(F) The cost is the same at 5 people but if there are more than five the group rate is better as we saw in part E. So the regular rate is better for less than 5 people.
(G) Here y = $150. Let us use the group rate formula and solve for x (the number of people). 10x+50 = 150 so 10x = 100 and x = 10. Since 10 is more than 5 this is the better deal. However if you don't believe it or want to double check we can solve for x using y = 150 and the regular rate equation. We get: 20x = 150 so x = 7.5 Since we can't bring half a person we would only be able to bring 7 and that is less than 10 so this is not the best choice. Use the group rate and bring 10 people!