Prepared by Muhammad Tayyab, Subject Specialist Mathematics, Govt Christian High School Daska
📌 Based on National Curriculum 2023 / PECTAA 2026 Syllabus
📖 What's Inside: This exercise covers real-life applications of quadratic equations – population growth, profit maximization, motion problems, and optimization. Perfect for Punjab Boards exam preparation.
📚 Related Resources – Unit 2: Quadratic Equations & Inequalities
1
A town's population is modeled by \(P(t) = -2t^2 + 40t + 800\), where \(t\) is years since 2020. Find the years when the population will be at least 1000.
Given modeled equation is
\[P(t) = -2t^2 + 40t + 800\]We have to find year when population will be at least 1000. So,
\[\begin{aligned} P(t) &\geq 1000 \\ -2t^2 + 40t + 800 &\geq 1000 \\ -2t^2 + 40t + 800 - 1000 &\geq 0 \\ -2t^2 + 40t - 200 &\geq 0 \\ -2(t^2 - 20t + 100) &\geq 0 \\ \Rightarrow t^2 - 20t + 100 &\leq 0 \end{aligned}\]Associated equation
\[t^2 - 20t + 100 = 0\]Here \(a = 1\), \(b = -20\), \(c = 100\)
\[\begin{aligned} t &= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\[6pt] t &= \frac{-(-20) \pm \sqrt{(-20)^2 - 4(1)(100)}}{2(1)} \\[6pt] t &= \frac{20 \pm \sqrt{400 - 400}}{2} \\[6pt] t &= \frac{20 \pm \sqrt{0}}{2} \\[6pt] t &= \frac{20 \pm 0}{2} \\[6pt] t &= \frac{20}{2} \\[4pt] t &= 10 \end{aligned}\]Since \(t = 10\) represents the number of years since 2020:
\[Year = 2020 + 10 = 2030\]The population will be at least 1000 in the year 2030.
2
A company models its profit \(P\) in thousands of rupees by \(P(x) = -5x^2 + 150x - 1000\), where \(x\) is the price per item in rupees. Find the price that gives maximum profit.
\[P(x) = -5x^2 + 150x - 1000\]
Here \(a = -5\), \(b = 150\), \(c = -1000\)
Since \(a = -5 < 0\), the parabola opens downward and the vertex gives the maximum point.
\[\begin{aligned} x &= -\frac{b}{2a} \\[6pt] x &= -\frac{150}{2(-5)} \\[6pt] x &= -\frac{150}{-10} \\[4pt] \boldsymbol{x} &= \boldsymbol{15} \end{aligned}\]The price per item that gives maximum profit is Rs. 15.
3
A toy car rolls down an incline and covers a distance given by \(d = t^2 - 0.5t\) metres, where \(t\) is the time in seconds. Find the time when the car has travelled a distance 12.5 metres.
\[\begin{aligned}
d &= t^2 - 0.5t \\
12.5 &= t^2 - 0.5t \quad \because d = 12.5 \\
0 &= t^2 - 0.5t - 12.5 \\
t^2 - 0.5t - 12.5 &= 0
\end{aligned}\]
Here \(a = 1\), \(b = -0.5\), \(c = -12.5\)
\[\begin{aligned} t &= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\[6pt] t &= \frac{-(-0.5) \pm \sqrt{(-0.5)^2 - 4(1)(-12.5)}}{2(1)} \\[6pt] t &= \frac{0.5 \pm \sqrt{0.25 + 50}}{2} \\[6pt] t &= \frac{0.5 \pm \sqrt{50.25}}{2} \\[6pt] t &= \frac{0.5 \pm 7.0887}{2} \end{aligned}\]either
\[\begin{aligned}
t &= \frac{0.5 + 7.0887}{2} \\[4pt]
t &= \frac{7.5887}{2} \\[4pt]
t &= 3.79 \text{ s}
\end{aligned}\]
or
\[\begin{aligned}
t &= \frac{0.5 - 7.0887}{2} \\[4pt]
t &= \frac{-6.5887}{2} \\[4pt]
t &= -3.29 \text{ s}
\end{aligned}\]
Rejected, time cannot be negative
The car will travel a distance of 12.5 metres after approximately 3.79 seconds.
4
A ball's height (in metres) after \(t\) seconds is \(h(t) = -4t^2 + 24t\). For what time interval is the ball at least 20 m above the ground?
Given
\[h(t) = -4t^2 + 24t\]According to given condition
\[\begin{aligned} h(t) &\geq 20 \\ -4t^2 + 24t &\geq 20 \\ -4t^2 + 24t - 20 &\geq 0 \\ -4(t^2 - 6t + 5) &\geq 0 \\ \Rightarrow t^2 - 6t + 5 &\leq 0 \end{aligned}\]Associated equation
\[\begin{aligned} t^2 - 6t + 5 &= 0 \\ t^2 - 5t - t + 5 &= 0 \\ t(t-5) - 1(t-5) &= 0 \\ (t-5)(t-1) &= 0 \end{aligned}\]either
\[\begin{aligned} t - 5 &= 0 \\ t &= 5 \end{aligned}\]
or
\[\begin{aligned} t - 1 &= 0 \\ t &= 1 \end{aligned}\]
Since \((t-5)(t-1) \leq 0\)
The ball is at least 20 m above the ground when \(\boldsymbol{1 \leq t \leq 5}\).
5
A ball is thrown upward with an initial velocity of \(40 \text{ ms}^{-1}\). Calculate the maximum height it reaches above ground level.
\(Initial\ velocity = v_i = 40\ \text{ms}^{-1}\)
\(Final\ Velocity = v_f = 0\ \text{ms}^{-1}\)
\(Gravitational\ acceleration = g = -10\ \text{ms}^{-2}\)
\(Height = h = ?\)
By using third equation of motion
\[\begin{aligned} 2aS &= v_f^2 - v_i^2 \\ 2gh &= v_f^2 - v_i^2 \\ 2(-10)(h) &= (0)^2 - (40)^2 \\ -20h &= 0 - 1600 \\ h &= \frac{-1600}{-20} \\[4pt] \boldsymbol{h} &= \boldsymbol{80 \text{ m}} \end{aligned}\]The maximum height reached by the ball is 80 metres.
6
A freelancer's earnings follow the model \(E(h) = -2h^2 + 40h\), where \(E\) is earning in rupees and \(h\) is hours worked per week. What is the maximum number of hours he should work to maximize earnings?
\[E(h) = -2h^2 + 40h + 0\]
Here \(a = -2\), \(b = 40\), \(c = 0\)
Since \(a = -2 < 0\), the parabola opens downward and the vertex gives the maximum point.
\[\begin{aligned} x &= -\frac{b}{2a} \\[6pt] x &= -\frac{40}{2(-2)} \\[6pt] x &= -\frac{40}{-4} \\[4pt] \boldsymbol{x} &= \boldsymbol{10} \end{aligned}\]He should work a maximum of 10 hours to maximize earnings.
📈 Key Concepts – Applications of Quadratics
- Optimization: For \(ax^2+bx+c\), vertex \(x = -\frac{b}{2a}\) gives maximum (if \(a<0\)) or minimum (if \(a>0\)).
- Inequalities: For "at least" or "at most" problems, set up and solve quadratic inequalities.
- Motion: Use equations of motion (\(v_f^2 = v_i^2 + 2as\)) for projectile problems.
- Real-world: Quadratic models appear in population, profit, distance, height, and optimization scenarios.