Overview
A downward-opening quadratic has a highest point, but a word problem may restrict the allowed inputs or change the model at a capacity boundary. The unrestricted vertex is then only a candidate. We will connect quadratic structure to height, demand and fencing, keeping the variable’s meaning and feasible domain visible throughout. A correct optimization must explain why the selected value is allowed and why no other allowed value is better.
Use the vertex when it lies inside the allowed interval
For 0 ≤ t ≤ 7, an object’s height is h(t) = −t² + 6t + 7. Complete the square to write h(t) = −(t − 3)² + 16. The squared term is nonnegative, so its negative is at most zero. The greatest height is therefore 16, attained when t = 3, which lies inside the stated interval.
The vertex form explains both the maximum and its location. The value 7 is the initial height, not the peak. Always check the domain after finding the vertex: if the time of the peak were outside the allowed interval, the best feasible value would have to be found at an allowed boundary. The physical interpretation does not disappear when an equation is supplied; it tells you which part of the parabola the problem actually permits you to use.
Set the requested event equal to the correct height
For h(t) = −2t² + 8t + 10, the initial height is h(0) = 10. Returning to that height means solving h(t) = 10, not h(t) = 0. This gives −2t² + 8t = 0, or −2t(t − 4) = 0. The roots are t = 0 and t = 4; the positive return time is four seconds.
The distractor two is the vertex time, answering when the height is greatest rather than when it returns. The value five corresponds to a different event involving height zero. Read the target event before solving. A height model can support questions about initial position, peak, return to initial position or contact with the ground, and each translates into a different equation even though the same formula appears in every case.
Respect a capacity boundary in a profit model
Demand falls from 80 items at $10 to 60 at $14, so it decreases by five items per dollar and equals 130 − 5p at price p. The shop can supply only fifty items, each costing $2. Until p = 16, demand exceeds capacity, so units sold stay at fifty and profit is 50(p − 2), increasing with price.
Above $16, demand is below capacity and profit becomes (p − 2)(130 − 5p). This downward quadratic has its unrestricted vertex at p = 14, but that value is outside this branch’s domain. On the allowed branch p ≥ 16, profit is already decreasing. The increasing and decreasing branches meet at p = 16, which is the maximizing price. The answer fourteen optimizes a demand-only formula where its assumed sales quantity would exceed the shop’s capacity.
Compare a fencing vertex with a width cap
A garden against a wall needs two end fences, one parallel fence and an additional divider perpendicular to the wall. If w is the distance from the wall and L the parallel length, forty meters of fencing gives 3w + L = 40. Thus area is w(40 − 3w) = 40w − 3w². Its unrestricted vertex is at w = 20/3, but the distance from the wall cannot exceed six meters.
The area increases throughout the allowed interval up to six, so use w = 6 and L = 22, giving 132 m². The option 400/3 ignores the cap. The value 168 omits the divider while keeping w = 6; 200 omits both the divider and the width cap. For transfer, define the variable, derive the model, identify every restriction and only then optimize. Check vertices, branch boundaries and feasible endpoints as appropriate. In practice, explain why the unconstrained answer is accepted or rejected rather than assuming every vertex is automatically the final solution.
Apply the method to a new case
A rectangle's perimeter is 20 meters, and its width must be at most 3 meters. Find the greatest possible area. If the width is w, the length is 10 − w and the area is A(w) = w(10 − w) = 25 − (w − 5)². The unconstrained maximum would occur at w = 5, but that violates w ≤ 3. For 0 < w ≤ 3, moving w toward 5 decreases the squared term, so area increases up to the allowed boundary w = 3. The length is 7 and the maximum area is 21 square meters. Reporting 25 ignores the binding constraint; reporting 10 answers a side-sum question. Check the perimeter, 2(3 + 7) = 20, and verify that a smaller allowed width such as 2 gives area 16. The model's vertex is only a candidate until the feasible domain is checked.