Answer to Question #261650 in Calculus for haemha

Question

 A vertical line passing through the point (1,2) intersects the X axis at A(a, 0) and Y axis at B(0, b). Find area of triangle of least area if a and b are positive.

Answer 1

Equation of AB ( intercept form)

$$\dfrac{x}{a}+\dfrac{y}{b}=1$$

A line passes through the point $(1,2)$

$$\dfrac{1}{a}+\dfrac{2}{b}=1$$

$$\dfrac{1}{a}=1-\dfrac{2}{b}$$

$$a=\dfrac{b}{b-2}, a>0, b>2$$

The area of $△AOB$ is

$$Area=\dfrac{ab}{2}$$

$$Area=A(b)=\dfrac{b^2}{2(b-2)}$$

Find the first derivative with respect to $b$

$$A'(b)=(\dfrac{b^2}{2(b-2)})'=\dfrac{2b(b-2)-b^2}{2(b-2)^2}=\dfrac{b^2-4b}{2(b-2)^2}$$

Find the critical number(s)

$$A'(b)=0=>\dfrac{b^2-4b}{2(b-2)^2}=0=>b_1=0, b_2=4$$

We consider $b>2.$

If $2<b<4,$ then $A'(b)<0,A(b)$ decreases.

If $b>4,$ then $A'(b)>0,A(b)$ increases.

The function $A(b)$ has a local minimum at $b=4.$

Since the function $A(b)$ has the only extremum for $b>2,$ then the function $A(b)$ has the absolute minimum for $b>2$ at $b=4.$

$$a=\dfrac{4}{4-2}=2$$

$$Area_{min}=\dfrac{2(2)}{2}=2(square\ units)$$