a Rectanglular pen for a pet is 5 feet longer than it is wide. give possible values for the width W of the pen if its area must be between 176 and 760 square feet inclusively. The width of the pen ranges from the smaller value of the width--- feet, to the larger value of the width , ----feet , inclusively.
w(w+5) = 176
w^2 + 5 w - 176 = 0
(w -11)(w+16) = 0
w = 11
w^2 + 5 w = 760
w^2 + 5 w - 760 = 0
use quadratic equation
w = 25.18, 30.18 or 25.18 for width
so
smaller is 11 and bigger is 25.18
To find the possible values for the width of the pen, we can set up an equation using the given information.
Let's denote the width of the pen as W (in feet). According to the problem, the length of the pen is 5 feet longer than its width, so the length would be W + 5.
The area of a rectangle is given by the formula: Area = Length × Width.
To find the possible values for the width, we need to solve the following inequality:
176 ≤ (W + 5) × W ≤ 760
First, let's simplify the inequality:
W × (W + 5) ≥ 176
W^2 + 5W ≥ 176
W × (W + 5) ≤ 760
W^2 + 5W ≤ 760
Now, we have two quadratic inequalities. We can solve them separately.
For the first inequality:
W^2 + 5W ≥ 176
We can solve this by subtracting 176 from both sides:
W^2 + 5W - 176 ≥ 0
Next, we can factorize the quadratic equation:
(W + 16)(W - 11) ≥ 0
Now, we have the factors. To find the possible values for W, we need to determine when the expression (W + 16)(W - 11) is greater than or equal to zero.
By analyzing the signs of the factors, we can create a sign chart:
-16 11
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Factor: + − +
(W + 16) (W - 11)
From the sign chart, we see that the expression is greater than or equal to zero when W is less than or equal to -16 or when W is greater than or equal to 11.
Now let's solve the second inequality:
W^2 + 5W ≤ 760
Subtract 760 from both sides:
W^2 + 5W - 760 ≤ 0
Factorize the quadratic equation:
(W + 40)(W - 19) ≤ 0
Again, use the signs of the factors to create a sign chart:
-40 19
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Factor: + − +
(W + 40) (W - 19)
From the sign chart, we see that the expression is less than or equal to zero when W is between -40 and 19, inclusively.
Combining the results from both inequalities, we find that the possible values for W are the intersection of the two solution sets:
W ≤ -16 or W ≥ 11 and -40 ≤ W ≤ 19
Therefore, the possible values for the width of the pen range from -40 feet to -16 feet, and from 11 feet to 19 feet inclusively.