Question 958223
perimeter  = 2 * length + 2 * width
set L = length , W = width
perimeter = 2L + 2W
since perimeter is to be 151
151 = 2L + 2W
Solving for L,
151 - 2W = 2L
dividing each side by2
{{{(151 - 2W)/2 = 2L/2}}}
{{{(151 - 2W)/2 = L}}}
Since area = L*W
{{{ area = ((151 - 2W)/2) * W }}}
{{{ area = (151W - 2W^2)/2 }}}
multiplying each side by 2
{{{ 2 * area = (151W - 2W^2) }}}
set y = 2*area
If we maximize the value of y in the equation
{{{ y = (151W - 2W^2) }}}
we will can determine the best dimensions for L and W
rewriting
{{{ y = -2W^2 + 151W }}}
Using the coordinate values a = -2 and b = 151
the value of W where y is maximum occurs when -b/2a is the value of W.
{{{-b/2a = -(151/2(-2))}}}
{{{-b/2a = 151/4}}}
area will be maximum when W = 151/4
We previously determined {{{ L = (151-2W)/2 }}} so substituting we get
{{{ L = (151-2(151/4))/2 }}}
{{{ L = (151-(151/2))/2 }}}
{{{ L = (151/2)/2 }}}
{{{ L = 151/4 }}}
So the maximum area will occur when W = 151/4 and L = 151/4