a satellite camera takes a rectangular-shaped picture. the smallest region that can be photographed is a 4km by 6km rectangle. as the camera zooms out, the length 1 and the width w of the rectangle increase at a rate of 3km/second. how long does it take for the area A to be at least 4 times its original size?
Possible Answers:A) 3.2 seconds B)1.61 seconds C) 4.94 seconds D) 9.7 seconds
Please help by showing how to put the "word problem" in a formula, so I can understand how you came to the correct answer.
To solve this problem, we need to find the relationship between the dimensions of the rectangle and its area, and then determine how long it takes for the area to be at least 4 times its original size.
Let's denote the original length of the rectangle as L₀ (given as 4 km) and the original width as W₀ (given as 6 km).
Since the length L increases at a rate of 3 km/second and the width W increases at a rate of 3 km/second, we can express the length and width as functions of time (t):
L = L₀ + 3t
W = W₀ + 3t
The area A of a rectangle is given by multiplying its length and width:
A = L * W
Substituting the expressions for L and W, we get:
A = (L₀ + 3t)(W₀ + 3t)
To find how long it takes for the area to be at least 4 times its original size, we need to solve the equation:
A ≥ 4A₀
Where A₀ is the original area (L₀ * W₀).
Substituting the expression for A and simplifying, we have:
(L₀ + 3t)(W₀ + 3t) ≥ 4(L₀ * W₀)
Expanding the left side of the inequality, we get:
L₀W₀ + 3L₀t + 3W₀t + 9t² ≥ 4L₀W₀
Rearranging the terms and simplifying, we obtain a quadratic inequality:
9t² + (3L₀ + 3W₀)t + (L₀W₀ - 4L₀W₀) ≥ 0
Simplifying further, we have:
9t² + (3L₀ + 3W₀)t - 3L₀W₀ ≥ 0
Now, we want to find the value of t when the left side of the inequality is equal to zero, as this represents the point where the area is exactly 4 times its original size.
To solve this quadratic inequality, we can calculate the discriminant (Δ) of the quadratic equation:
Δ = (3L₀ + 3W₀)² - 4(9)(-3L₀W₀)
Evaluating the discriminant further, we have:
Δ = 9(L₀ + W₀)² + 108L₀W₀
Since we want to find when the quadratic is equal to zero, we set Δ = 0 and solve for t:
0 = 9(L₀ + W₀)² + 108L₀W₀
Simplifying the equation and factoring, we get:
0 = 9(L₀ + W₀)² + 12L₀W₀(L₀ + W₀)
Dividing by 9 and simplifying further, we have:
(L₀ + W₀)² + 4L₀W₀(L₀ + W₀) = 0
Factoring out (L₀ + W₀), we obtain:
(L₀ + W₀)[(L₀ + W₀) + 4L₀W₀] = 0
Now we have two possible cases:
1) L₀ + W₀ = 0
2) (L₀ + W₀) + 4L₀W₀ = 0
Case 1: L₀ + W₀ = 0
If L₀ + W₀ = 0, it means that the original length and width were negative or zero, which is not possible in this case. Therefore, we disregard this case.
Case 2: (L₀ + W₀) + 4L₀W₀ = 0
Simplifying the equation, we have:
(L₀ + W₀)(1 + 4L₀W₀) = 0
Again we have two possibilities:
a) L₀ + W₀ = 0
b) 1 + 4L₀W₀ = 0
a) L₀ + W₀ = 0
If L₀ + W₀ = 0, it means that the current length and width are negative or zero, which is not possible. Therefore, we disregard this case.
b) 1 + 4L₀W₀ = 0
Solving for L₀W₀, we get:
L₀W₀ = -1/4
Since area can't be negative, we disregard this case as well.
From the above analysis, we see that there are no values of t that satisfy the condition for the area to be at least 4 times its original size.
Therefore, the answer is none of the given options (A, B, C, or D).