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

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

Solutions

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

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

Therefore from right angled  Δ ABC ,

We have 
  AB  = AC cosθ          ∵ AC = 2r
Let A(x) be the area of Rectangle ABCD
∴ A(x) = AB × BC
    A = (2r cos θ)(2r sin θ )
    A =  4r² sin θ cos θ
    A = 2r²  (2sin θ cos θ)
    A = 2r²  (sin 2θ )

⇒ 2r² 2 (cos 2θ ) = 0 ,As r² is constant
⇒cos 2θ = 0
⇒cos 2θ =cos (π/2)
⇒ θ = π/4

 =4r²  (-2sin 2θ ) 

∴ A has Maximum value at θ = π/4

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

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) =   x³ + (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’  .

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