A rectangular pen for a pet is 5 feet longer than it is wide. Give possible width of the pen if its area must be 204 and 750 square feet, inclusively.
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its area must be 204 and 750 square feet, inclusively?
perimeter = 2(width+length)
length = width+5
a = w(w+5)
so, given the limits on the area,
204 <= w(w+5) <= 750
if w(w+5) >= 204,
w^2 + 5w - 204 >= 0
w>=12
if w(w+5) <= 750
w^2+5w-750 <= 0
w<=25
so, 12 <= w <= 25
To find the possible width of the pen, we can set up an equation based on the given information.
Let's assume the width of the pen is "x" feet.
According to the given information, the length of the pen is 5 feet longer than its width, so the length would be "x + 5" feet.
The area of a rectangle is calculated by multiplying its length and width. Therefore, the equation for the area of the pen is:
Area = length × width
Plugging in the given areas:
204 = (x + 5)x
750 = (x + 5)x
Now we can solve these equations to find the possible values of x.
For the first equation (204 = (x + 5)x), we can rearrange it and solve for x:
x^2 + 5x - 204 = 0
We can solve this quadratic equation by factoring or using the quadratic formula. Factoring the equation, we have:
(x + 17)(x - 12) = 0
Setting each factor equal to zero:
x + 17 = 0 or x - 12 = 0
Solving for x:
x = -17 or x = 12
Since a width cannot be negative, we disregard x = -17. Therefore, one possible width of the pen is x = 12 feet.
For the second equation (750 = (x + 5)x), we again rearrange it and solve for x:
x^2 + 5x - 750 = 0
Using factoring or the quadratic formula, we find:
(x + 30)(x - 25) = 0
Setting each factor equal to zero:
x + 30 = 0 or x - 25 = 0
Solving for x:
x = -30 or x = 25
Again, we disregard x = -30 since the width cannot be negative, leaving us with x = 25 feet as a possible width of the pen.
In conclusion, the possible widths of the rectangular pen are 12 feet and 25 feet.