Break-Even With an Unattainable Non-Integer Solution
A couple compares two reception venues for a wedding with a guest list that is not final. Venue A charges a $500 room fee plus $18 per guest. Venue B charges a $1,200 room fee plus $10 per guest, and both quotes include food, staff, and tax. A student sets 500 + 18g = 1,200 + 10g, solves to get g = 87.5, and writes in the recommendation, 'The venues cost the same for 87.5 guests, so Venue B is the better deal for our wedding.' The couple expects between 80 and 120 guests and must sign a contract with an exact guaranteed count. The student's algebra is correct. The tutor must decide whether the recommendation follows from it, how to handle a non-integer solution in a counting context, and how to present the comparison so the couple can act on it. Venue B also requires a minimum of 100 guests.
1. Solve the system and interpret the solution, including what happens at 87 guests and at 88 guests.
Read the worked answer
500 + 18g = 1,200 + 10g gives 8g = 700, so g = 87.5 and the common cost is 500 + 18(87.5) = $2,075 (check: 1,200 + 10(87.5) = 2,075).
Because guests are counted in whole people, g = 87.5 cannot occur. Evaluate on either side:
- g = 87: A = 500 + 1,566 = $2,066; B = 1,200 + 870 = $2,070. A is cheaper by $4.
- g = 88: A = 500 + 1,584 = $2,084; B = 1,200 + 880 = $2,080. B is cheaper by $4.
The intersection is a genuine break-even in the algebraic sense and the crossover point in the economic sense, but no achievable contract produces equal costs. The sign of B - A = 700 - 8g flips across g = 87.5: negative (B cheaper) above it, positive (A cheaper) below it.
2. Evaluate the student's recommendation that Venue B is the better deal, and write the advice the couple actually needs.
Read the worked answer
The recommendation does not follow. Equality at 87.5 guests says nothing about which venue is cheaper at any count the couple can actually contract; it says only where the two cost functions cross.
At 87 guests A saves $4, at 88 guests B saves $4, and the gap widens by $8 per guest as the count moves away from 87.5. At the top of the expected range, 120 guests: A = 500 + 2,160 = $2,660; B = 1,200 + 1,200 = $2,400, so B saves $260. At 80 guests: A = $1,940; B = $2,000, so A saves $60.
The advice should be a decision rule keyed to the final guaranteed count: choose A if the count is 87 or fewer, B if it is 88 or more, and note that the choice is close (under $10) only in the 86-89 range. Because the couple's uncertainty spans the crossover, the practical recommendation is to pin down the count before signing, not to pick a venue on the basis of a fractional break-even.
3. Venue B turns out to require a minimum of 100 guests. Explain how this constraint changes the analysis and whether the break-even point still matters.
Read the worked answer
The constraint does not change the algebra: the cost functions and their intersection at g = 87.5 are unaffected. It changes the feasible region. Contracts with B are now possible only for g ≥ 100 (and, if the couple's maximum holds, g ≤ 120).
On that restricted interval, B - A = 700 - 8g is negative throughout, e.g., at g = 100, B = $2,200 and A = $2,300, so B is cheaper by $100. The break-even at 87.5 lies outside the achievable region, so it is mathematically valid but operationally irrelevant: within every contract the couple can sign, B is the lower-cost choice.
This is the judgment the exam is testing: a solution to an equation must be checked against the constraints of the situation. The tutor should have the student restate the answer as 'for 100 to 120 guests, B costs $700 - 8g less than A,' rather than quoting the unreachable crossover as a decision point.