Quadratic Optimization

Quadratic functions model countless real-world situations: the revenue from selling a product, the area of a fenced region, the height of a thrown ball, the cost of manufacturing. Because every quadratic has a single vertex — a highest or lowest point — we can find the optimal value of any quantity modeled by a quadratic. This is quadratic optimization.

Why the Vertex Is the Key

A quadratic function $f(x) = ax^2 + bx + c$ has a parabola as its graph. The vertex of that parabola is the maximum or minimum point:

• If $a > 0$ (opens upward), the vertex is the minimum
• If $a < 0$ (opens downward), the vertex is the maximum

The $x$-coordinate of the vertex is:

$x = -\frac{b}{2a}$

The optimal value (the $y$-coordinate) is $f\!\left(-\frac{b}{2a}\right)$.

This formula comes from completing the square, which rewrites $ax^2 + bx + c$ as:

$a\!\left(x + \frac{b}{2a}\right)^{\!2} + c - \frac{b^2}{4a}$

The squared term is minimized (equals zero) when $x = -\frac{b}{2a}$.

Completing the Square to Find the Vertex

Worked Example 1: Standard Form to Vertex Form

Convert $f(x) = 2x^2 - 12x + 7$ to vertex form.

Step 1: Factor out the leading coefficient from the first two terms:

$f(x) = 2(x^2 - 6x) + 7$

Step 2: Complete the square inside the parentheses. Half of $-6$ is $-3$; square it to get $9$:

$f(x) = 2(x^2 - 6x + 9 - 9) + 7 = 2(x - 3)^2 - 18 + 7 = 2(x - 3)^2 - 11$

Vertex: $(3, -11)$. Since $a = 2 > 0$, this is a minimum.

Quick Formula Method

Using $x = -\frac{b}{2a} = -\frac{-12}{2(2)} = \frac{12}{4} = 3$:

$f(3) = 2(9) - 12(3) + 7 = 18 - 36 + 7 = -11$

Both methods confirm the vertex is $(3, -11)$.

Optimization Word Problems

The general strategy for quadratic optimization problems:

1. Define variables and write the quantity to optimize as a function
2. Express everything in one variable using the given constraint
3. Find the vertex using $x = -\frac{b}{2a}$ or completing the square
4. Interpret the result in context and check that it makes sense

Worked Example 2: Maximum Revenue

A company sells widgets at a price of $p$ dollars each. Market research shows that the number of units sold is $n = 800 - 20p$. Find the price that maximizes revenue.

Step 1: Revenue = (price)(quantity): $R(p) = p \cdot (800 - 20p) = 800p - 20p^2$.

Step 2: This is a quadratic in $p$ with $a = -20$ (opens downward), so it has a maximum.

Step 3: $p = -\frac{b}{2a} = -\frac{800}{2(-20)} = -\frac{800}{-40} = 20$

Step 4: Maximum revenue: $R(20) = 800(20) - 20(400) = 16000 - 8000 = 8000$.

Answer: The optimal price is $20 per widget, producing a maximum revenue of $8,000. At this price, $n = 800 - 20(20) = 400$ widgets are sold.

Worked Example 3: Maximum Area with Fixed Perimeter (Carpentry)

A carpenter has 60 feet of trim to frame a rectangular window. What dimensions maximize the area?

Step 1: Let $x$ = width. Then the perimeter constraint gives: $2x + 2y = 60$, so $y = 30 - x$.

Step 2: Area: $A(x) = x(30 - x) = 30x - x^2$.

Step 3: $x = -\frac{30}{2(-1)} = 15$ feet, so $y = 30 - 15 = 15$ feet.

Step 4: Maximum area: $A(15) = 15(15) = 225$ square feet.

Answer: A square window (15 ft by 15 ft) maximizes the area at 225 sq ft. This confirms the classic result: among all rectangles with a given perimeter, the square has the greatest area.

Worked Example 4: Minimum Cost (Engineering)

A manufacturer’s cost function is $C(x) = 0.01x^2 - 8x + 3000$, where $x$ is the number of units produced daily.

Step 1: Here $a = 0.01 > 0$, so the parabola opens upward — the vertex is a minimum.

Step 2: $x = -\frac{-8}{2(0.01)} = \frac{8}{0.02} = 400$ units.

Step 3: Minimum cost: $C(400) = 0.01(160000) - 8(400) + 3000 = 1600 - 3200 + 3000 = 1400$.

Answer: Producing 400 units daily minimizes the cost at $1,400.

Worked Example 5: Three Sides of Fencing

A farmer needs to fence a rectangular area along a river (no fence needed on the river side). With 200 meters of fencing, what dimensions maximize the enclosed area?

Step 1: Let $x$ = the side perpendicular to the river (two of these). Then the side parallel to the river is $200 - 2x$.

Step 2: Area: $A(x) = x(200 - 2x) = 200x - 2x^2$.

Step 3: $x = -\frac{200}{2(-2)} = 50$ meters. The parallel side is $200 - 2(50) = 100$ meters.

Step 4: Maximum area: $A(50) = 50(100) = 5000$ square meters.

Answer: The optimal pen is 50 m by 100 m, enclosing 5,000 sq m.

Worked Example 6: Projectile Motion

A ball is launched upward from a 48-foot platform with an initial velocity of 64 ft/s. Its height is $h(t) = -16t^2 + 64t + 48$.

Maximum height time: $t = -\frac{64}{2(-16)} = \frac{64}{32} = 2$ seconds.

Maximum height: $h(2) = -16(4) + 64(2) + 48 = -64 + 128 + 48 = 112$ feet.

When does it hit the ground? Set $h(t) = 0$:

$-16t^2 + 64t + 48 = 0 \implies t^2 - 4t - 3 = 0$

$t = \frac{4 \pm \sqrt{16 + 12}}{2} = \frac{4 \pm \sqrt{28}}{2} = 2 \pm \sqrt{7} \approx 2 \pm 2.646$

Taking the positive root: $t \approx 4.646$ seconds.

Second Derivative Test Preview

In calculus, you will learn that the second derivative of $f(x) = ax^2 + bx + c$ is $f''(x) = 2a$. This constant tells you the concavity:

• $f''(x) = 2a > 0$: concave up, so the critical point is a minimum
• $f''(x) = 2a < 0$: concave down, so the critical point is a maximum

This is exactly what the sign of $a$ already tells you in the quadratic case, but the second derivative test generalizes to all differentiable functions — not just quadratics.

Common Mistakes

1. Forgetting domain restrictions. If $x$ represents a physical quantity (like length), it must be positive. Always check that your optimal $x$-value is in the feasible domain.

2. Confusing $x$ and $y$ in the answer. The problem asks for dimensions, not just the vertex coordinates. Translate back to the original variables.

3. Not checking the sign of $a$. The vertex is a maximum only when $a < 0$, and a minimum only when $a > 0$.

Practice Problems

Test your understanding with these problems. Click to reveal each answer.

Key Takeaways

• The vertex of a quadratic $f(x) = ax^2 + bx + c$ occurs at $x = -\frac{b}{2a}$
• If $a > 0$, the vertex is a minimum; if $a < 0$, the vertex is a maximum
• For optimization problems: define the objective function, use constraints to reduce to one variable, then find the vertex
• Among rectangles with a fixed perimeter, the square has the maximum area
• Among rectangles with three sides fenced, the optimal width is half the length
• Always check that the optimal solution lies within the physically meaningful domain
• The second derivative test ($f''(x) = 2a$) extends the vertex analysis to calculus

Return to College Algebra for more topics in this section.