## Find two positive numbers whose sum is 16 and sum of whose cube is Minimum

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**Show that of all the rectangles inscribed in a circle of given radius . The Square has maximum Area.**

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**Solutions**

**Let ABCD be rectangle which is inscribed in a given circle of radius ‘r’**

And Let Î¸ be the angle between side of rectangle and Diameter of given circle.

Therefore from right angled Î” ABC ,

We have

**AB = AC cos**Î¸

**∵ AC = 2r**

Let A(x) be the area of Rectangle ABCD

∴ A(x) = AB × BC

A = (2r cos Î¸)(2r sin Î¸ )

A = 4r

^{2 }sin Î¸ cos Î¸

A = 2r

^{2}(2sin Î¸ cos Î¸)

A = 2r

^{2}(sin 2Î¸ )

⇒ 2r

^{2}2 (cos 2Î¸ ) = 0 ,As r

^{2}is constant

⇒cos 2Î¸ = 0

⇒cos 2Î¸ =cos (Ï€/2)

⇒ Î¸ = Ï€/4

∴ A has Maximum value at Î¸ = Ï€/4

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**Find two positive numbers whose sum is 16 and sum of whose cube is Minimum**

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**Solution**

Let us consider two numbers x and 16- x .Then transforming our problem to mathematical form which says “sum of whose cube” as follows

A (x) =

^{3}+ (16 - x)

^{3}…….. (1)

Differentiating both sides w .r. t “x” , we get

^{ }

X = 8

So x = 8 will be the 1

^{st}required numbers if Double derivatives of A w. r. t ‘x’ comes to be positive at x = 8.

Differentiate (2) w. r. t. ‘x’ .

## My previous Posts

How to find area of the circles which is interior to the parabola

**How to find common area of two parabolas.**##

FINAL WORDS

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