# Show that of all the rectangles inscribed in a given fixed circle, the square has the maximum area

**Solution:**

Maxima and minima are known as the extrema of a function.

Maxima and minima are the maximum or the minimum value of a function within the given set of ranges.

Let a rectangle of length l and breadth b be inscribed in the given circle of radius a.

Then, the diagonal passes through the centre and is of length 2a cm.

Now, by applying the Pythagoras theorem, we have:

(2a)^{2} = l ^{2} + b^{2}

⇒ b^{2} = 4a^{2} - l^{2}

⇒ b = √4a² - l²

A =l √4a² - l²

Therefore,

dA/dl = √4a² - l² + l (1/2√4a² - l²)(- 2l)

= √4a² - l² - l²/(√4a² - l²)

= 4a^{2} - l^{2}/√4a² - l²

d^{2}A/dl^{2}

= [√4a² - l² (- 4l) - (4a^{2} - 2l^{2}) (- 2l) / (2√4a² - l²))] / (4a^{2} - l^{2})

= (- 12a^{2}l + 2l^{3})/(4a^{2} - l^{2})^{3/2}

= (- 2l (6a^{2} - l^{2}))/(4a^{2} - l^{2})^{3/2}

Now,

dA/dl = 0

Hence,

⇒ (4a^{2} - 2l^{2}) / (√4a² - l²) = 0

⇒ 4a^{2} = 2l ^{2}

⇒ l = √2 a

Thus,

b = √4a² - 2a²

= √2a²

= √2 a

When, l = √2 a

Then,

d^{2}A/dl^{2} = - 2 (√2 a)(6a^{2} - 2a^{2}) / (2√2 a^{3})

= (- 8√2 a^{3})/(2√2 a^{3})

= - 4 < 0

By the second derivative test, when l = √2 a then the area of the rectangle is the maximum.

Since,

l = b = √2 a,

the rectangle is a square.

Hence, it has been proved that of all the rectangles inscribed in the given fixed circle, the square has the maximum area

NCERT Solutions Class 12 Maths - Chapter 6 Exercise 6.5 Question 19

## Show that of all the rectangles inscribed in a given fixed circle, the square has the maximum area

**Summary:**

Hence we have shown that of all the rectangles inscribed in a given fixed circle, the square has the maximum area. Maxima and minima are known as the extrema of a function

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