Answer:
Step-by-step explanation:
The area is given in two ways: 1) as a formula: A = 30w - w^2, and 2) as a specific numerical value: A = 200 m^2.
Equating these, we get:
A = 200 m^2 = A = 30w - w^2
Rewriting this equation in the standard form of a quadratic function:
-30w + w^2 + 200 m^2 = 0
or w^2 - 30w + 200 = 0, which in factored form is (w - 10)(w - 20) = 0.
Then w = 10 and w = 20.
The perimeter of the rectangle is P = 60 m, and this equals 2w + 2l. Therefore, 30 m = w + l, or l = 30 - w.
We have already found that w could be either 10 or 20.
If w = 10, then l = 30 - 10 = 20, and the perimeter would thus be:
P = 2(10) + 2(20) = 20 + 40 = 60.
This satisfies the constraints on w.
The width of the rectangle is 10 meters. The length is 20 meters.
Based on the information, the width of the rectangle will be 10 meters.
Based on the information given, the equation to solve the question will be:
30w - w² = 200
-w² + 30w - 200 = 0
w² - 30w + 200 = 0
w² - 10w - 20w + 200 = 0
w(w - 10) - 20(w - 10)
(w - 10) (w - 20) = 0
w - 10 = 0
w = 0 + 10
w = 10
Therefore, the width of the rectangle is 10 meters.
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