The area a, in square meters, of a rectangle with a perimeter of 60 meters is given by the equation a = 30w − w2, where w is the width of the rectangle in meters. what is the width of a rectangle if its area is 200 m2? m

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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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.

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your answer would be 10x - 1