Q.(a) Divide a number 15 into two parts such that the square of one part multiplied with the cube of the other part is maximum.

Maximization With Constraint: The Art of Doing Your Best With Limits

Imagine you are a student with ₹100 to spend on lunch. You love pizza (₹20 a slice) and you love juice (₹10 a glass). You want the most satisfying meal — the maximum "happiness" — but you cannot spend more than ₹100. That is a constrained maximization problem.

Without the ₹100 limit, you would eat infinite pizza and drink infinite juice. But reality imposes a boundary. The question becomes: given your limited budget, which combination of pizza and juice gives you the greatest possible satisfaction?

This is the core of every constrained maximization problem: you have something you want to maximize (happiness, profit, area, speed) and a restriction that limits your choices (money, materials, time, a physical law). The trick is to find the best possible outcome within that restriction, not the best possible outcome in an imaginary world with no limits.

The Precise Statement

A constrained maximization problem has three parts:

1. The Objective Function — the thing you want to maximize. Call it $f(x, y)$. In our example, $f$ is your "satisfaction" from $x$ slices of pizza and $y$ glasses of juice.

2. The Constraint — the rule that limits your choices. Usually written as $g(x, y) = c$ (or $\leq c$). Here: $20x + 10y = 100$ (you must spend exactly ₹100).

3. The Decision Variables — the things you control. Here: $x$ (pizza slices) and $y$ (glasses of juice).

The problem is:

You are searching over all pairs $(x, y)$ that satisfy the constraint, and picking the one that makes $f$ largest.

The Key Insight: Tangency

Here is the conceptual leap. Imagine you draw a map of your satisfaction — contour lines where $f$ is constant (like elevation lines on a hiking map). Each contour is a curve: "all combinations of pizza and juice that give 10 units of happiness," then 11, then 12, and so on.

Now draw the budget line: $20x + 10y = 100$, a straight line.

You want the highest possible contour that still touches the budget line. That happens where the budget line is tangent to a contour — just barely kissing it. If the budget line cuts through a contour, you can slide along it to a higher contour. The optimum is the point where you cannot improve without leaving the budget.

The Method: Lagrange Multipliers

The tangency condition leads to a systematic method. To maximize $f(x, y)$ subject to $g(x, y) = c$:

1. Form the Lagrangian:

2. Set all partial derivatives to zero:

3. Solve the system. The third equation just gives back $g(x, y) = c$. The first two enforce $\nabla f = \lambda \nabla g$. …