Question

plzz say me the answer​

Answer 1

Ans is option A c^3/6√3....

Let the length, breadth and height of the box be l,x,and y respectively.

Area = c^2sq. units.

∴x^2+4xy=c^2

y=(c^2−x^2)/4x

Let v be the volume of the box, then

v=x^2y

⇒v=x^2(c^2−x^2)/4x

v=c^2/4x−x^3/4

Differentiate w.r.t x we get,

dv/dx=c^2/4−3x^2/4

Again differentiate w.r.t x we get,

d^2v/dx^2=−3x/2

For maximum or minimum , we must have,

dv/dx=0⇒c^2/4−3x^2/4=0

⇒3x^2/4=c^2/4

⇒x=c/√3

(d^2v/dx^2) at x=c/√3 gives -3c/2√3 <0

Thus, v is maximum when x=c/√3

Put x=c/√3 we get

y=c/2√3

∴ The maximum volume of the box is given by

v=x^2y

=c^2 /3 × c/2√3

=c^3/6√3

Answer 2

Answer: The reason is the nucleophile attack at ortho and pera position of benzene only.

I was unable to comment there so I posted here.