Answer to Question #261650 in Calculus for haemha

Question #261650

 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.


1
Expert's answer
2021-11-08T08:12:00-0500

Equation of AB ( intercept form)


xa+yb=1\dfrac{x}{a}+\dfrac{y}{b}=1

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


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

1a=12b\dfrac{1}{a}=1-\dfrac{2}{b}

a=bb2,a>0,b>2a=\dfrac{b}{b-2}, a>0, b>2

The area of AOB△AOB is


Area=ab2Area=\dfrac{ab}{2}

Area=A(b)=b22(b2)Area=A(b)=\dfrac{b^2}{2(b-2)}

Find the first derivative with respect to bb


A(b)=(b22(b2))=2b(b2)b22(b2)2=b24b2(b2)2A'(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=>b24b2(b2)2=0=>b1=0,b2=4A'(b)=0=>\dfrac{b^2-4b}{2(b-2)^2}=0=>b_1=0, b_2=4

We consider b>2.b>2.

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


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

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

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


a=442=2a=\dfrac{4}{4-2}=2

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


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