Q.Case Study -2 LED bulbs are energy-efficient because they use significantly less electricity than traditional bulbs while producing the same amount of light. They convert more energy into light rather than heat, reducing waste. Additionally, their long lifespan means fewer replacements, saving resources and money over time. A company manufactures a new type of energy -efficient LED bulb. The cost of production and the revenue generated by selling x bulbs (in an hour) are modelled as 𝐶(𝑥) = 0.5𝑥2 − 10𝑥 + 150 and 𝑅(𝑥) = −0.3𝑥2 + 20𝑥 respectively, where 𝐶(𝑥) and 𝑅(𝑥) are both in ₹. To maximize the profit, the company needs to analyze these functions using calculus. Use the given models to answer the following questions: I. Derive the profit function 𝑃(𝑥) [1] II. Find the critical points of 𝑃(𝑥). [1] III
(A) Determine whether the critical points correspond to a maximum or a minimum profit by using the second derivative test.

Profit Maximization

A business earns revenue from selling goods and pays out costs to make them. Profit is simply what is left over:

where $x$ is the number of units produced (or sold), $R(x)$ is the total revenue and $C(x)$ is the total cost. Profit maximization asks a very concrete question: what value of $x$ makes $P(x)$ as large as possible? In the Application of Derivatives chapter this is just a maxima-and-minima problem in disguise.

The intuition

Produce very little and you earn almost nothing; produce far too much and your costs run ahead of your revenue. Somewhere in between there is a sweet spot. As you increase production, profit first rises, then flattens, then falls — so at the top the rate of change of profit is momentarily zero. That is exactly what a derivative detects.

Finding the maximum

Treat $P(x)$ as a differentiable function and follow the standard optimisation steps:

1. Differentiate: $P'(x) = R'(x) - C'(x)$.
2. Set the derivative to zero to locate critical points: $P'(x) = 0 \ \Rightarrow\ R'(x) = C'(x)$.
3. Confirm it is a maximum with the second-derivative test: profit is maximised at a critical point $x^{*}$ where

A quick illustration

Suppose $P(x) = 100x - 2x^2$ (in rupees). Then $P'(x) = 100 - 4x$. Setting $P'(x) = 0$ gives $x = 25$. Since $P''(x) = -4 < 0$, the profit is indeed greatest at $x = 25$ units. …

Optimization Word Problems

Imagine planning a garden with 40 metres of fencing and wanting the largest rectangular area. A long, thin rectangle wastes space; a square feels roomier; somewhere in between lies the best shape. That is an optimisation problem — a fixed resource and a quantity to make as large (or as small) as possible.

Every optimisation word problem has the same skeleton: the best outcome — maximum area, minimum cost, largest volume, shortest time — under a constraint — limited material, a fixed budget, a given perimeter.

The Plan of Attack

The problem gives you a story, not a graph. Your job is to turn it into a single-variable function and find its peak or valley:

1. Name the quantity to optimise — call it $Q$, and write it using variables.
2. Find the constraint — a relation between those variables (e.g. "perimeter $=40$").
3. Reduce to one variable — use the constraint to eliminate the rest.
4. Differentiate — solve $Q'(x)=0$ to find the critical points.
5. Confirm — use $Q''(x)<0$ for a maximum or $Q''(x)>0$ for a minimum.
6. Answer the question asked — give the actual dimensions/cost, not just $x$.

The Garden, Worked

40 m of fencing encloses a rectangle; maximise the area.

• Objective: $A = l\,w$.
• Constraint: $2l+2w=40$, so $l+w=20$.
• Reduce: $w=20-l$, giving $A(l)=l(20-l)=20l-l^2$.
• Differentiate: $A'(l)=20-2l=0 \implies l=10$.
• Confirm: $A''(l)=-2<0$, a maximum.

So $l=w=10$ m — a 10 m $\times$ 10 m square.

The Common Families

| Problem type | Typical objective | Typical constraint |

|--------------|-------------------|--------------------| …