You can put this solution on YOUR website! 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
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
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:
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.
When a quadratic is in standard form, , the x-coordinate of the vertex is . So we will put our equation into standard form:
with the "w" playing the role of "x" and the "A" being the "y". Now we can find the w-coordinate of the vertex:
which simplifies to:
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:
Simplifying...
So the maximum possible area is 529/4 square meters.
You can put this solution on YOUR website! a square is the rectangle with maximum area.
all 4 sides in a square are equal
46/4=11.5
11.5^2=132.25 or as the other tutor says 529/4
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