What is the area of the largest rectangle that can be inscribed in a semicircle of radius 10 units?

A rectangle is inscribed in a semi circle with radius $r$ with one of its sides at the diameter of the semi circle. Find the dimensions of the rectangle so that its area is a maximum.

My Try:

Let length of the side be $x$, Then the length of the other side is $2\sqrt{r^2 -x^2}$, as shown in the image.

Then the area function is

$$A(x) = 2x\sqrt{r^2-x^2}$$

$$\begin{align}A'(x) &= 2\sqrt{r^2-x^2}-\frac{4x}{\sqrt{r^2-x^2}}\\ &=\frac{2}{\sqrt{r^2-x^2}} (r^2 - 2x -x^2)\end{align}$$

setting $A'(x) = 0$,

$$\implies x^2 +2x -r^2 = 0$$

Solving, I obtained:

$$x = -1 \pm \sqrt{1+r^2}$$

That however is not the correct answer, I cannot see where I've gone wrong? Can someone point out any errors and guide me the correct direction. I have a feeling that I have erred in the differentiation.

Also how do I show that area obtained is a maximum, because the double derivative test here is long and tedious.

Thanks!

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Given a semicircle of radius r, we have to find the largest rectangle that can be inscribed in the semicircle, with base lying on the diameter.
Examples:

Input : r = 4 Output : 16 Input : r = 5 Output :25

Let r be the radius of the semicircle, x one half of the base of the rectangle, and y the height of the rectangle. We want to maximize the area, A = 2xy. So from the diagram we have,

y = √(r^2 – x^2)

So, A = 2*x*(√(r^2 – x^2)), or dA/dx = 2*√(r^2 – x^2) -2*x^2/√(r^2 – x^2) Setting this derivative equal to 0 and solving for x,

dA/dx = 0

or, 2*√(r^2 – x^2) – 2*x^2/√(r^2 – x^2) = 0
2r^2 – 4x^2 = 0
x = r/√2This is the maximum of the area as,

dA/dx > 0 when x > r/√2

and, dA/dx < 0 when x > r/√2
Since y =√(r^2 – x^2) we then have
y = r/√2
Thus, the base of the rectangle has length = r/√2 and its height has length √2*r/2.
So, Area, A=r^2

#include <bits/stdc++.h>
using namespace std;
float rectanglearea(float r)
{
    if (r < 0)
        return -1;
    float a = r * r;
    return a;
}
int main()
{
    float r = 5;
    cout << rectanglearea(r) << endl;
    return 0;
}

class GFG
{
static float rectanglearea(float r)
{
if (r < 0)
    return -1;
float a = r * r;



return a;
}
public static void main(String[] args)
{
    float r = 5;
    System.out.println((int)rectanglearea(r));
}
}

def rectanglearea(r) :
    if r < 0 :
        return -1
    a = r * r
    return a
if __name__ == "__main__" :
    r = 5
    print(rectanglearea(r))

using System;
class GFG
{
static float rectanglearea(float r)
{
if (r < 0)



    return -1;
float a = r * r;
return a;
}
public static void Main()
{
    float r = 5;
    Console.Write((int)rectanglearea(r));
}
}

<?php
function rectanglearea($r)
{
    if ($r < 0)
        return -1;
    $a = $r * $r;
    return $a;
}
$r = 5;
echo rectanglearea($r)."\n";
?>

function rectanglearea(r)

{

if (r < 0)

return -1;

var a = r * r;

return a;

}

var r = 5;

document.write(parseInt(rectanglearea(r)));

</script>

OUTPUT :

Time Complexity: O(1)
Auxiliary Space: O(1)

Given a semicircle of radius , find the largest rectangle (in terms of volume) that can be inscribed in the semicircle, with base lying on the diameter.

Let be the radius of the semicircle, one half of the base of the rectangle, and the height of the rectangle. We want to maximize the area,

. Referencing the diagram we have

Thus,

Setting this derivative equal to 0 and solving for ,

This is a maximum of the area since

Since

we then have

Thus, the base of the rectangle has length

and its height has length