A fish tank in the shape of a cuboid has length 320 cm. Its length is twice that of its width. To enhance viewing, the area of the four vertical faces should be maximized.
Find the optimum "viewing area" of a fish tank that is fixed to the wall so that the area of three faces only should be considered.
Length - 2x
width - x
hight - h
But now I'm so confused. Plese could you be so nice and help me ? Thank you so much
Don't know if you paraphrased the question, but the question is indeed confusing.
First it gave the length of 320 cm and width of 320/2=160 cm. So there is no variable to optimized. Then it talks about maximizing 4 faces, and then jumped to optimizing 3 faces.
Finally, it gives length, width and height as 2x, x and h, without any constraints.
You cannot do an optimizing problem without constraints, since you can increase h without limit to maximize the area.
Please check the question for typos or missed information.
REMEMBER: Check you question for accuracy before posting. This will help you get your answer with less exchanges.
Thank you for your advice, I appreciate it really a lot, but unfortunately I always checked me writing and exercise is exactly copied from the book Oxvord IB diploma programme Mathematical studies standard level chapter 4 page 183. Why I was very confused because I'm student from internationl school and thought that I don't exactly understand English very well. But luckily I figured out whole exercise.
Anyway really thank you for your time. Have a very nice day.
How did you figure it out?
Sure, I can help you with that!
To find the optimum viewing area of the fish tank, we need to maximize the area of the three vertical faces. Let's break it down step by step:
1. Let's start by setting up the given information. We are told that the length of the tank is 320 cm and that it is twice the width. So we can express the dimensions as follows:
Length (L) = 320 cm
Width (W) = L/2 = 320/2 = 160 cm
2. Now, let's define the height of the fish tank as 'H'. However, since the height is not specified in the problem, we can leave it as a variable for now.
3. The area of a single vertical face of the cuboid is given by the formula:
Face Area = Width * Height
4. Now, let's calculate the total viewing area of the three faces. Remember, we are excluding the base of the fish tank from consideration:
Total Viewing Area (A) = Face Area + Face Area + Face Area
= 3 * (Width * Height)
= 3 * (160 cm * H)
= 480H cm^2
5. To maximize the viewing area, we need to find the optimal height (H) that will give us the maximum value for the total viewing area.
6. To optimize this problem, we can take the derivative of the viewing area formula with respect to H and set it equal to zero to find the maximum:
dA/dH = 480 cm^2
7. Solving for H, we get:
480 = 0
Since this equation has no solution, it means that the viewing area increases indefinitely as the height becomes larger.
8. However, in practical terms, there will be limitations on the maximum height. So, in this case, we can just choose a reasonable height that still provides a good viewing area.
Therefore, the optimum viewing area of the fish tank is infinite, but in practice, you can choose a reasonable height that suits your needs and space limitations.