Q.Find the maximum area of an isosceles triangle inscribed in the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with its vertex at one end of the major axis.

Setup: Vertex $V=(a,0)$; base vertices $(x,y),(x,-y)$ on the ellipse, $y>0$.

Area: $A=y(a-x)$, and from the ellipse $y^2=\dfrac{b^2}{a^2}(a^2-x^2)$.

Maximising $Z=A^2=\dfrac{b^2}{a^2}(a-x)^3(a+x)$ (single variable $x$):

$\frac{dZ}{dx}=-\frac{2b^2}{a^2}(a-x)^2(a+2x)=0 \implies x=-\frac a2 \ \text{(rejecting }x=a\text{, which gives zero area)}$ …

Setting up the geometry

The vertex of the isosceles triangle is at one end of the major axis, $V=(a,0)$. Since the triangle is isosceles with its vertex on the axis of symmetry, the other two vertices must be placed symmetrically about the $x$-axis: $P=(x,y)$ and $Q=(x,-y)$, with $y>0$ and $-a<x<a$, both lying on the ellipse.

Writing the area as a function of $x$ alone

The base $PQ$ has length $2y$ and lies on the vertical line through $x$; the height from this base to the vertex $V=(a,0)$ is $a-x$.

Since $(x,y)$ lies on the ellipse,

so $y=\dfrac{b}{a}\sqrt{a^2-x^2}$ (taking the positive root since $y>0$). Substituting,

This is now a function of the single variable $x$ — exactly the kind of one-variable optimisation this chapter covers.

Differentiating

Since $A(x)\ge 0$ on this domain, it is simpler to maximise $Z(x)=A(x)^2$ (a maximum of $A$ occurs at the same $x$ as a maximum of $A^2$, because squaring is an increasing operation for non-negative numbers).

Differentiate using the product rule:

Solving $\dfrac{dZ}{dx}=0$

$(a-x)^2=0 \implies x=a \quad\text{(rejected — this makes y=0, i.e. zero area)}$ …

Method: Lagrange Multipliers for a Geometric Area-Maximisation Problem

This method solves "maximise/minimise an area (or other quantity) subject to points lying on a given curve" problems — the general technique whenever an optimisation has two variables tied together by a constraint equation such as an ellipse or circle.

Steps

Step 1: Set up coordinates using the symmetry of the figure

Place the curve in standard position and use any reflective symmetry to describe the unknown points with as few variables as possible (e.g. an isosceles shape symmetric about an axis needs only one pair $(x,y)$ for the non-fixed vertices, since the mirror point is $(x,-y)$).

Step 2: Write the objective function $f$ and the constraint $g=0$

Express the quantity to optimise (area, using $\tfrac12\times\text{base}\times\text{height}$ here) purely in terms of the chosen coordinates, and write the curve's equation in the form $g(x,y)=0$.

Step 3: Form $\nabla f = \lambda \nabla g$

Compute both partial derivatives of $f$ and $g$ with respect to each variable, and set up the two resulting equations relating them through $\lambda$.

$\frac{\partial f}{\partial x} = \lambda \frac{\partial g}{\partial x}, \qquad \frac{\partial f}{\partial y} = \lambda \frac{\partial g}{\partial y}$ …

Common Mistakes

Mistake 1: Losing a variable's sign or dividing by a coordinate that could be zero

Why it's wrong: eliminating $\lambda$ often requires dividing by $x$ or $y$; if that variable could legitimately be zero, that root gets silently thrown away without ever being checked as a separate case. Correct approach: before dividing, note explicitly that the variable must be nonzero for a non-degenerate figure, rather than dividing silently.

Mistake 2: Keeping every root produced by the constraint quadratic without checking which is geometrically valid …