Find the dimensions of a rectangle with perimeter 76 m whose area is as large as possible.?

1 Answer

• Well,

  this proof does not use derivatives

  let a = 76/4 = 19m

  then the square of side a=19 has the area

  A = 19^2 = 381

  let e >= 0 represent a variation

  let the length of any rectangle of perimeter p = 76m be :

  L = a + e/2

  and

  let the width of any rectangle of perimeter p = 76m be :

  l = a – e/2

  then we verify that the total perimeter is :

  2L + 2l = 2(a + e/2) + 2(a – e/2)

  = 2a + e + 2a – e

  = 2a

  = p = 76m

  and the area of this rectangle is a function of e, so:

  A(e) = (a+ e/2) * (a – e/2)

  = a^2 – e^2/4

  is less, whenever e > 0 (meaning e >=0 AND e =/=0 !!)

  in other words, the area of any rectangle of perimeter p is less then the Maximum reached for the square of side :

  a = p/4 = 19m

  conclusion :

  the largest area is reached for the square of side 76/4 = 19m

  qed

  hope it’ ll help !!