Quadratic Functions & Applications: Optimization, Sketching, Revenue–Cost Models

Scenario
- Brick wall provides one side; contractor can supply only 100 m of fencing material for the other three sides.
- Variables
- L = length (side opposite the wall)
- W = width (two identical sides perpendicular to the wall)
Formulas
- Area of enclosure: A = W \times L
- Available fencing gives a linear constraint: 2W + L = 100
Expressing W in terms of L
- 2W = 100 - L \;\Rightarrow\; W = 50 - \frac{L}{2}
Area as a single‐variable quadratic
- A(L) = \Bigl(50 - \frac{L}{2}\Bigr) L = 50L - \frac{L^{2}}{2}
Domain
- Physical lengths imply L \ge 0.
- Area becomes negative when L > 100 (see parabola crossing the A=0 axis), so practical domain: (0 , 100).
Graph
- Parabolic curve opening downward (negative quadratic term).
- Vertex gives the maximum attainable area (not explicitly computed here, but concept foreshadows later vertex method).
- Links to upcoming quadratic‐function theory.

General form: f(x) = ax^{2} + bx + c with a \ne 0.
Coefficients
- a = leading coefficient (controls opening direction & width).
- b, c = remaining linear & constant terms.
Example (simplest): f(x) = x^{2}
- Here a = 1,\; b = 0,\; c = 0.
- Point table: x=-2,-1,0,1,2 \;\Rightarrow\; f(x)=4,1,0,1,4.
- Graph features
- Vertex at (0,0).
- Axis of symmetry x=0.
- Concave upward (because a>0).

Concavity
- a>0 ⇒ opens up (cup).
- a<0 ⇒ opens down (cap).
y–intercept
- Set x=0 ⇒ point (0,c).
x–intercepts (roots)
- Solve ax^{2}+bx+c=0.
- Quadratic formula: x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}.
Vertex
- x_{v} = -\dfrac{b}{2a};
- y_{v} = f\bigl(-\dfrac{b}{2a}\bigr).
Axis of symmetry
- Vertical line x = -\dfrac{b}{2a} (passes through vertex, mirrors graph).

Identify coefficients: a=-4,\; b=-8,\; c=5.
Step 1 (concavity): a<0 ⇒ concave down.
Step 2 (y–int): (0,5).
Step 3 (x–ints):
- Discriminant b^{2}-4ac = 64 - 4(-4)(5)=64+80=144.
- Roots x = \dfrac{8 \pm 12}{-8} \Rightarrow -\tfrac{5}{2},\; \tfrac{1}{2}.
Step 4 (vertex):
- x_{v} = -\tfrac{b}{2a}= \tfrac{8}{-8} = -1.
- y_{v} = f(-1)= -4(1)+8+5 = 9 ⇒ vertex (-1,9).
Step 5 (axis): x = -1.
Graph passes all plotted checkpoints; calculator verification confirms.

Revenue: R(x)= -x^{2}+40x
Cost: C(x)=8x+192
Break-even (solve R=C)
- -x^{2}+40x = 8x+192 \;\Rightarrow\; x^{2}-32x+192=0
- Factor: (x-8)(x-24)=0 \Rightarrow x=8,\,24.
- Minimum break-even quantity: 8 units.
Maximum revenue
- Vertex x_{v}= -\dfrac{b}{2a}=20.
- R(20)= -400+800 = 400 ⇒ max revenue $400.
Profit function
- P(x) = R - C = -x^{2}+32x-192 (also quadratic, a<0).
- Vertex x_{v}=16 ⇒ P(16)=64.
- Max profit $64, achieved at 16 units.

Demand equation: Q = -0.06P + 84
- Q = members, P = annual membership fee ($).
Revenue:
- R(P) = P \times Q = -0.06P^{2} + 84P (concave down).
Cost structure:
- Fixed: 20{,}000 per year.
- Variable: 20 per member ⇒ 20Q.
- Substitute demand into cost:
  C(P) = 20{,}000 + 20(-0.06P + 84)= -1.2P + 21{,}680.
Profit:
- \Pi(P) = R(P) - C(P)= -0.06P^{2}+85.2P -21{,}680.
Max revenue
- Vertex of R: P = -\dfrac{84}{2(-0.06)} = 700.
- R(700)= -0.06(700)^{2}+84(700)= 29{,}400.
Max profit
- Vertex of \Pi: P = -\dfrac{85.2}{2(-0.06)} = 710.
- \Pi(710)= -0.06(710)^{2}+85.2(710)-21{,}680 = 8{,}566.
Interpretation
- Pricing at $700 maximizes revenue but not necessarily profit.
- Slightly higher price $710 yields lower membership count but maximizes profit.
- Illustrates trade-off between volume and margin—core managerial insight.

Quadratics naturally model area optimization, projectile motion, break-even analysis, and revenue–cost–profit relationships.
The vertex is the key for extrema (max/min) due to parabolic symmetry.
Domain restrictions (physical or business) must be checked alongside algebraic solutions.
Profit differs from revenue; optimal decisions may shift when costs are included.
Five-step sketch method provides a quick qualitative understanding before exact computations or software graphing.
Quadratic formula remains an indispensable tool for roots, informing intercepts and feasibility boundaries.