Question 395574
There are multiple ways to prove that the maximum area is a square. You could let A = w*(152 - w), then take the derivative or find the vertex, as the other tutor did.


I like the classic AM-GM inequality: AM-GM (arithmetic mean - geometric mean inequality) says that for positive real numbers {{{a[i]}}}, the arithmetic mean of all {{{a[i]}}}'s is greater than or equal to the geometric mean. If we let w and 152-w be the sides,


{{{(w + (152-w))/2 >= sqrt(w*(152-w))}}}


{{{ 76 >= sqrt(w*(152-w))}}}


{{{76^2 >= w*(152-w)}}} --> the area is less than or equal to 76^2. AM-GM says that the equality case occurs if and only if all the {{{a[i]}}}'s are equal, or w = 304-w --> w = 76, area = 76^2.


This is a slightly unconventional way, but it's the easiest way to prove the similar problem that says that the largest possible volume of a rectangular solid of fixed surface area occurs when the solid is a cube.