This problem is bugging me:
a rectangle storage container with an open top is to have a volume of 10m3. the length of the base is twice its width. material for the base costs $10 per square meter. material for the sides cost $6 per square meter. find the cost of the cheapest container.
2006-10-10 13:56:52 · 4 answers · asked by pheonixfire2018 1 in Science & Mathematics ➔ Mathematics
Actually, this is a calculus problem. Here are the steps:
1. Call the dimensions of the base x and 2x. (It's twice as long as it is wide.) Call the height y.
2. Write the formula for the volume and set it equal to 10 m^3.
3. Solve for y in terms of x. Now you know the length, width, and height in terms of x. (They are 2x, x, and whatever you got when you solved for y.)
4. Now write an expression for the cost of the materials. It is the area of the base times 10 (dollars per m^2), plus the area of the sides times 6. (Hint: The area of the sides is the perimeter of the base multiplied by the height.)
5. Now the calculus: Take the derivative of the cost of materials with respect to x. Take that expression (the derivative) and set it equal to zero. Solve for x. (This will probably involve solving a quadratic, so you will have more than one solution. You'll have to figure out which solution is meaningful (i.e., x and y are greater than 0) and produces a minimum cost (as opposed to a "relative maximum" cost).)
Follow the steps and you should be able to do it.
Good luck.
2006-10-10 14:20:33 · answer #1 · answered by actuator 5 · 0⤊ 0⤋
As in line with my adventure you are able to shrink fee via utilising least width you have 3 dimensions chosen width as least it incredibly is going to provide least overlaps length could be midsection vale of your measurement top may be the optimal measurement eg in case you may desire to make container of 215 X a hundred and seventy X 80 minimum fee configuration would be L= a hundred and seventy W=80 H = 215 desire it helps
2016-12-13 06:00:47 · answer #2 · answered by Anonymous · 0⤊ 0⤋
Base = 4.5^(1/3) x (twice that amount) = 1.65m*3.3m = 5.451m^2
Cost of base = $54.51
Height = 10/5.451 = 1.834m
Area of sides = 1.834(2*1.65 + 2*3.3) = 18.1566m^2
Cost of sides = 6*18.1566 = $108.94
Total cost = 108.94 + 54.51 = $163.45
2006-10-10 14:28:54 · answer #3 · answered by Steve 7 · 0⤊ 0⤋
reminds me of quadratic stuffs,
i know how to do it but am just lazy to. we did it in grade ten
2006-10-10 14:04:34 · answer #4 · answered by Anonymous · 0⤊ 0⤋