Arectangle is inscribed in an equilateral triangle so that one side of the rectangle lies on the base of the triangle. find the maximum area the rectangle can have when the triangle has side length 14 inches.

A(max)  =  42.43 in²

Dimensions:

a   =  7 in

b   =  6,06  in

Step-by-step explanation:  See annex

Equilateral triangle  side  L  = 14 in,  internal angles all equal to 60°

Let  A  area of rectangle     A  = a*b

side   b        tan∠60°  = √3    tan∠60° = b/x     b  =  √3  * x

side  a   a = L  - 2x      a =  14  - 2x

A(x) = a*b        A(x) = ( 14  -  2x  ) *  √3  * x

A(x) =  14*√3*x  - 2√3 * x²

Taking derivatives both sides of the equation

A´(x)  =  14√3  -  4√3*x

A´(x)  = 0     ⇒    14√3  -  4√3*x  =  0     ⇒ 14  - 4x  = 0    x = 14/4

x  = 3,5  in

Then  

a =   14  - 2x          a = 14 - 7       a =  7 in

b  = √3*3,5               b = *√3 *3,5      b  = 6,06 in

A(max)  =  7 *6,06  

A(max)  =  42.43 in²

i think the answer is b)

step-by-step explanation: