Please explain in full.
Find the dimensions of a rectangle with perimeter 76 m whose area is as large as possible. (If both values are the same number, enter it into both blanks.)
m (smaller value)
m (larger value)
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 !!