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

Fence–Wall Optimization Problem

• 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.

Quadratic Functions: Definition & Terminology

• 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).

Five-Step Quick Sketch Method for Any Quadratic

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

Worked Sketch Example: $f(x) = -4x^{2} - 8x + 5$

• 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.

Business Application 1: Break-Even & Profit with Simple Quadratics

• 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.

Business Application 2: Health-Club Pricing (Demand, Revenue, Cost, Profit)

• 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.

Conceptual & Practical Takeaways

• 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.