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.

Question

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

Answer 1

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.

Answer 2

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.

Answer 3

How did you figure it out?