Question 828224
The general formula for perimeter of a rectangle is:
P = 2l + 2w
where "l" is the length and "w" is the width. For our rectangle the equation is:
46 = 2l + 2w<br>
Now we will take a moment to solve our perimeter equation for l. Subtracting 2w we get:
46 - 2w = 2l
Dividing by 2 we get:
23 - w = l<br>
The general formula for area of a rectangle is:
A = l*w
We solving the perimeter equation for l so we could substitute in for l in the area formula:
A = (23 - w)*w
Simplifying we get:
{{{A = 23w - w^2}}}<br>
The graph of this equation will be a parabola which opens downward (because of the "-" in front of the squared term). Its maximum area will be at the vertex of this parabola. So our task is to find the vertex of the parabola.<br>
When a quadratic is in standard form, {{{y = ax^2+bx+c}}}, the x-coordinate of the vertex is {{{(-b)/2a}}}. So we will put our equation into standard form:
{{{A = -w^2 + 23w}}}
with the "w" playing the role of "x" and the "A" being the "y". Now we can find the w-coordinate of the vertex:
{{{w[v]=(-23)/2(-1)}}}
which simplifies to:
{{{w[v]=23/2}}}
This is the width that creates the maximum area. It is not the maximum area. For that we have to put 23/2 in for the w in:
{{{A = -w^2 + 23w}}}
{{{A = -(23/2)^2 + 23(23/2)}}}
Simplifying...
{{{A = -(23/2)^2 + 23(23/2)}}}
{{{A = -(529/4) + (529/2)}}}
{{{A = -(529/4) + (1058/4)}}}
{{{A = 529/4}}}
So the maximum possible area is 529/4 square meters.