Q.Find two positive numbers whose sum is $15$ and the sum of whose squares is minimum.

Idea: Use the constraint to write everything in one variable, then minimise with a single derivative (constraint-substitution, no Lagrange).

Let the two positive numbers be $x$ and $15-x$, with $0<x<15$. Minimise the sum of squares

Differentiate and set to zero:

$S'(x)=4x-30=0\ \Rightarrow\ x=\tfrac{15}{2}=7.5.$ …

Set up in one variable

Let the two positive numbers be $x$ and $15-x$ (their sum is $15$), where $0<x<15$. The quantity to minimise is the sum of their squares:

Expanding,

Because we used the constraint $x+(15-x)=15$ to eliminate the second number, this is now a single-variable function — ordinary calculus finishes the job.

Minimise

Differentiate:

Set $S'(x)=0$:

Check the nature of this point:

so $S$ is minimised (the graph is an upward parabola). The value $x=7.5$ is positive and less than $15$, so it is valid.

Answer the question …

Method: Reducing a Constrained Optimization Problem to One Variable

Any problem of the form "given a fixed total, split it to optimise some combined quantity" follows this same recipe.

Steps

Step 1: Name the unknowns and write the constraint

If two (or more) quantities are linked by a fixed sum, product, or other relation, write that relation down explicitly first — it is what lets you eliminate a variable.

Step 2: Use the constraint to express everything in one variable

Solve the constraint for one unknown in terms of the other, and substitute it into the quantity you're optimising, so the objective becomes a function of a single variable.

Step 3: Differentiate and solve for critical points

Also note the valid range for the variable coming from the problem's own conditions (for example, "positive numbers" restricts the domain). …