Maximizing the Area of a Triangle in the First Quadrant

[babbagee]

Here is the question

You are planning to close off a corner of the first quadrant with a line segment 20units long running from (a,0) to (0,b). Show that the area of the triangle enclosed by the segment is largest when a = b.

 

[himanshu121]

AREA=1/2*a*b

And also Apply the Distance formula to find the relation b/w a and b eliminate one from the two equations and use calculus to find the max of AREA

 

[HallsofIvy]

Or use "Lagrange multiplier" method:

To maximize (1/2)ab subject to the requirement that x/a+ y/b= 1 (the equation of the line from (a,0) to (0,b)) we must have
The vector (1/2)b i+ (1/2)a j (the grad of (1/2)ab) parallel to the vector (1/a)i+ (1/b)j (the grad of x/a+ y/b) (in an "ab" coordinate system of course).
That is (1/2)b= λ(1/a) and (1/2)a= λ(1/b) where λ is the Lagrange multiplier. Dividing the first equation by the second to eliminate λ, b/a= a/b or a²= b² so a= b or a= -b. Since this is in the first quadrant, a= b.