An aquarium is designing a new exhibit to showcase tropical fish. The exhibit will include a tank that is a rectangular prism with a length, $l$ , that is twice the width, $w$ . The volume of the tank is $420$ cubic feet. What is the width of the tank to the nearest tenth of a foot? $\newline$ Hint: There is another way to represent the length of the tank in terms other than $l$ .

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Q. An aquarium is designing a new exhibit to showcase tropical fish. The exhibit will include a tank that is a rectangular prism with a length,

l

, that is twice the width,

w

. The volume of the tank is

420

cubic feet. What is the width of the tank to the nearest tenth of a foot?

\newline

Hint: There is another way to represent the length of the tank in terms other than

l

Given Information: We are given that the volume of the tank is $420$ cubic feet and that the length is twice the width. Let's denote the width as $w$ and the length as $2w$ . The height of the tank is not given, so we'll denote it as $h$ . The volume of a rectangular prism is given by the formula $V = lwh$ , where $l$ is the length, $w$ is the width, and $h$ is the height.
Volume Equation: Since we know the volume $V$ is $420$ cubic feet, we can write the equation $420 = (2w) \cdot w \cdot h$ , which simplifies to $420 = 2w^2 \cdot h$ .
Height in Terms of Width: We need to find the width $w$ , but we have two variables in the equation. However, we can express the height $h$ in terms of the width $w$ and the volume $V$ . Let's rearrange the equation to solve for $h$ : $h = \frac{420}{2w^2}$ .
Finding Width: We don't have enough information to find the exact value of $h$ , but we don't need it to find the width ( $w$ ). We can assume that the height ( $h$ ) is such that it will give us a whole number for the width ( $w$ ) when we solve the equation. This is because the problem does not provide any information about the height, and we are only asked to find the width.
Factors of $420$ : To find the width $w$ , we need to find the value of $w$ that satisfies the equation $420 = 2w^2 \times h$ . Since we don't have the value of $h$ , we can't solve this equation directly. However, we can look for factors of $420$ that are perfect squares, as $2w^2$ must be a factor of $420$ for the equation to hold true.
Checking Factors: The factors of $420$ that are perfect squares are $1$ , $4$ , and $16$ . We can divide $420$ by each of these factors and see if the resulting quotient is an even number, which would correspond to $2w^2$ .
Identifying Correct Factor: Dividing $420$ by $4$ , we get $105$ , which is not an even number, so $4$ is not the correct factor. Dividing $420$ by $16$ , we get $26.25$ , which is not an even number either. Therefore, $16$ is not the correct factor. However, when we divide $420$ by $1$ , we get $420$ , which is an even number. This suggests that $4$ $1$ could be $420$ .
Solving for Width: If $2w^2$ is $420$ , then $w^2$ is $210$ . We can now take the square root of both sides to solve for $w$ : $w = \sqrt{210}$ .
Calculating Width: Calculating the square root of $210$ , we get $w \approx 14.49$ feet. Rounding to the nearest tenth of a foot, we get $w \approx 14.5$ feet.