Question

A rectangle is constructed with its base on the diameter of a semicircle with radius 23 and with its two other vertices on the semicircle. What are the dimensions of the rectangle with maximum​ area?

Answer 1

Answer:

$L=23\sqrt{2}$

$B=\frac{23}{\sqrt{2}}$      

The area of the rectangle is 529 unit square.        

Step-by-step explanation:

Given : A rectangle is constructed with its base on the diameter of a semicircle with radius 23 and with its two other vertices on the semicircle.

To find : What are the dimensions of the rectangle with maximum​ area?

Solution :

Let origin be the center of the circle with radius 23.

The equation of the semicircle is $x^2+y^2=23^2$

$x^2+y^2=529$

Let (z,0) and (-z,0) be the points on the diameter of the semicircle.  

∴ $z^2+y^2=529$

$y^2=529-z^2$

$y=\pm\sqrt{529-z^2}$

Since the semicircle in on y-axis so coordinates of point on the semicircle are

$(z,\sqrt{529-z^2})(-z,\sqrt{529-z^2})$

So, The length of the rectangle is $L=2z$

Breadth of the rectangle is  $B=\sqrt{529-z^2}$

Area of the rectangle is $A=L\times B$

$A=2z\times\sqrt{529-z^2}$

To maximum we have to derivate w.r.t z to find critical point.

$\frac{dA}{dz}=2(\sqrt{529-z^2})+2z\times \frac{1}{2}\times (-2z)\times(\frac{1}{\sqrt{529-z^2}})$

$0=2(\sqrt{529-z^2})-\frac{2z^2}{\sqrt{529-z^2}}$

$\frac{2(529-z^2)-2z^2}{\sqrt{529-z^2}}=0$

$1058-4z^2=0$

$4z^2=1058$

$z^2=\frac{529}{2}$

$z=\sqrt{\frac{529}{2}}$

$z=\frac{23}{\sqrt{2}}$

Substitute the value of z in the area and dimensions.

Length is $L=2z$

$L=2\times \frac{23}{\sqrt{2}}$

$L=23\sqrt{2}$

Breadth is $B=\sqrt{529-z^2}$

$B=\sqrt{529-\frac{529}{2}}$

$B=\sqrt{\frac{1058-529}{2}}$

$B=\sqrt{\frac{529}{2}}$

$B=\frac{23}{\sqrt{2}}$

Area is $A=23\sqrt{2}\times \frac{23}{\sqrt{2}}$

$A=23\times23$

$A=529$

Therefore, The area of the rectangle is 529 unit square.