So I am having a bit of trouble with this question. I get that I have to use optimisation but I am not sure how. 'find the least area of sheet metal required to make an open baking dish of square base and vertical sides capacity 2048cm cubed. I have tried to use optimisation by making the lenght times width times height formula = 2048, but I don't think that is what I should be doing. It is confusing me because there are so many unknown varliables.
Asked
Active
Viewed 21 times
0
-
Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or closed. To prevent that, please [edit] the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. – José Carlos Santos Oct 24 '20 at 06:57
-
@JoséCarlosSantos I am really sorry, what do I need to add for it to be ok? – lily Oct 24 '20 at 07:02
-
Tell us what you've tried. – José Carlos Santos Oct 24 '20 at 07:03
1 Answers
1
It is open container so there are $5$ surfaces. Now say the side of the square base is $x$ and height of the container is $y$
Given volume $V = x^2y = 2048$ ...(i).
You need to minimize the surface area as you need to use least amount of metal sheet.
Total surface area $S = x^2 + 4xy$ ...(ii)
$S = x^2 + \frac{2048 \times 4}{x}$ (substituting value of $y$ from (i))
At extrema, $\frac{dS}{dx} = 2x - \frac{2048 \times 4}{x^2} = 0 \implies x^3 = 4096$
This gives you $x = 16$ (side of the square base) and from (i), you can find the height which comes to $8$.
Now you can find minimum $S$ using (ii) (you can also do second derivative test to make sure this is minima, which it is).
Math Lover
- 51,819