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.
Sure! Let's break down the problem and solve it step by step.
Let's assume that after t seconds, the length of the rectangle is 1 + 3t km, and the width of the rectangle is w km.
The formula for the area of a rectangle is A = length * width. Substituting the given values, we have A = (1 + 3t) * w.
We're given that the original area (A₀) is equal to a 4 km by 6 km rectangle, so A₀ = 4 km * 6 km = 24 km².
We need to find how long it takes for the area to be at least 4 times its original size. So, we need to find the value of t when A ≥ 4A₀ (where A₀ is the original area).
Substituting the values, we get (1 + 3t) * w ≥ 4 * 24.
Simplifying the equation, we have (1 + 3t) * w ≥ 96.
Now, let's solve for t:
w ≥ 96 / (1 + 3t).
To find the minimum value for t, we need to find the maximum value for w. We know that w is increasing at a rate of 3 km/second. So, the maximum value for w is when t = 0 (initially).
Substituting t = 0 into the equation, we get w ≥ 96 / (1 + 3 * 0) = 96.
Therefore, the smallest value for w is 96 km, and this occurs when t = 0.
Now, let's find the time it takes for A to be at least 4 times its original size:
4A₀ = 4 * 24 = 96.
Substituting into the equation, we have w ≥ 96 / (1 + 3t).
Plugging in the values, we get 96 ≥ 96 / (1 + 3t).
Canceling out the 96 on both sides, we have 1 ≥ 1 / (1 + 3t).
Cross-multiplying, we get 1 + 3t ≥ 1.
Simplifying, we have 3t ≥ 0.
Dividing both sides by 3, we get t ≥ 0.
Since t represents time, it cannot be negative. Therefore, t = 0, and it takes 0 seconds for the area to be at least 4 times its original size.
So, the correct answer is none of the given options.
Please note that this is a simplified version of the problem, and other factors like the rate of change of w may affect the answer in a more realistic scenario.
To solve this problem, we first need to set up an equation for the area of the rectangle, A, as a function of time, t.
Given that the length increases at a rate of 3km/second, the length of the rectangle can be expressed as 4 + 3t (since the smallest region that can be photographed is a 4km by 6km rectangle).
Similarly, the width w of the rectangle can be expressed as w = 6 + 3t (since it increases at a rate of 3km/second).
The area A of the rectangle is given by the product of its length and width, so we have:
A = (4 + 3t)(6 + 3t)
Now, we need to find the time t for which the area A is at least 4 times its original size.
We can also express the original area A0 as (4)(6) = 24 square kilometers.
To find the time t, we can set up the following inequality:
A ≥ 4A0
(4 + 3t)(6 + 3t) ≥ 4(24)
Now, we can solve this inequality to find the value of t when the area A is at least 4 times its original size.
(4 + 3t)(6 + 3t) ≥ 96
Expanding the left side of the inequality:
24 + 18t + 12t + 9t^2 ≥ 96
9t^2 + 30t + 24 ≥ 96
Rearranging the equation:
9t^2 + 30t - 72 ≥ 0
Now, we can solve this quadratic inequality. We can factor the equation:
3(3t - 6)(t + 4) ≥ 0
Considering that the factors 3(3t - 6) and (t + 4) represent lengths, they must be greater than or equal to zero (≥ 0) to satisfy the conditions of the problem.
Case 1: 3t - 6 ≥ 0
3t ≥ 6
t ≥ 2
Case 2: t + 4 ≥ 0
t ≥ -4
Since time cannot be negative, we discard the second case t ≥ -4.
Therefore, t ≥ 2.
So it takes at least 2 seconds for the area A to be at least 4 times its original size.
Now we need to evaluate the answer choices given:
A) 3.2 seconds - Not the correct answer.
B) 1.61 seconds - Not the correct answer.
C) 4.94 seconds - Not the correct answer.
D) 9.7 seconds - Not the correct answer.
Since none of the answer choices satisfy the requirement of t ≥ 2, the answer is not given.