To contain the greatest area, the rectangle must be 10 m × 10 m.
Step-by-step explanation:
Let be the length and let be the length of the rectangular lot. It follows that its perimeter is equal to
Divide both sides by 2 and solve for , we yield
Recall the formula for the area of a rectangle: Area = length times width. In this case, the length is and the width is .
Using the formula, the area is equal to
To find the maximum value of the area, we need to treat the area as a function. Express the area in terms of a function as
The graph of this function is a downward parabola, and in particular, the maximum will occur at the vertex of the parabola.
In order to find its vertex, we should convert the function into the vertex form f(x) = a(x-h)² + k where (h, k) is the vertex.
The vertex is (10, 100). Therefore, the maximum value of this function is 100, and it is also the greatest possible area of the rectangle. This rectangle has dimensions 10 × 10. We must include the unit, in this case, the unit is meters.
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Answer:
To contain the greatest area, the rectangle must be 10 m × 10 m.
Step-by-step explanation:
Let
be the length and let
be the length of the rectangular lot. It follows that its perimeter is equal to
Divide both sides by 2 and solve for
, we yield
Recall the formula for the area of a rectangle: Area = length times width. In this case, the length is
and the width is
.
Using the formula, the area is equal to
To find the maximum value of the area, we need to treat the area as a function. Express the area in terms of a function as
The graph of this function is a downward parabola, and in particular, the maximum will occur at the vertex of the parabola.
In order to find its vertex, we should convert the function into the vertex form f(x) = a(x-h)² + k where (h, k) is the vertex.
The vertex is (10, 100). Therefore, the maximum value of this function is 100, and it is also the greatest possible area of the rectangle. This rectangle has dimensions 10 × 10. We must include the unit, in this case, the unit is meters.