Question

show that the rectangleof maximum area that can be inscribed in a circle of radius r is a square of side √2r​

Answer 1

Step-by-step explanation:

Please first see the image attached with this answer.

Let length of rectangle be x

Let breadth of rectangle be y

Let radius of circle be r

Step 1:

By pythagoras theorem:

                      x² + y² = 4r²                 (given)

                              y² = 4r² - x²

                              y   = (4r² - x²)½

=>   Area of rectangle(A) = xy ________1

Now put value of y in equation 1

=>                       A = x × (4r² - x²)½

=>                       Differentiate w.r.t. x

=>       da/dx = (4r² - x²)½ + xd(4r² - x²)½ /dx

=>  da/dx = (4r² - x²)½ + x(-2x)/2(4r² - x²)½

=>  da/dx = -x² + 4r² - x²/(4r² - x²)½

=>  da/dx = -2x² + 4r²/(4r² - x²)½

=>  da/dx = 4r² - 2x²/(4r² - x²)½ __________2

Now, we need to put da/dx=0 for min. or max. value of x in terms of r

=>  0 = 4r² - 2x²/(4r² - x²)½

=>  0 = 4r² - 2x²

=> 2x² = 4r²

=> x² = 2r²

=> x = (2)½r

Step 2:

Differentiate equation 2 once more to proof if value is max or min.

if d²a/dx² comes negative then value is max and if d²a/dx² comes positive then values is min.

=>d²a/dx² =(4r² - x²)½ (-4x)-(4r²-2x²)((-2x)/2(4r² - x²)½ )/4r² - x²

=> now put x = (2)½r

after simplifying we get,

=>   d²a/dx² = -8r²/2r²

=> d²a/dx² = -4 => negative value

=> since value of d²a/dx² is negative therefore this value is maximum when

x = (2)½r

=> therefore area is maximum at x = (2)½r

Hence proved

hope you understand the answer.Concept of maxima minma is bit hard to grasp but practicing the question makes it easy and its guaranteed 6 marker question in 12th boards.

Regards:

Aakash Maurya