Optimization Problem and Graphing Problem!!!

codyski95

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Optimization Problem

A rectangular storage container with an open top is to have a volume of 30 cubic meters. The length of its base is twice the width. Material for the base costs 11 dollars per square meter. Material for the sides costs 5 dollars per square meter. Find the cost of materials for the cheapest such container. Answer needs to be rounded to the nearest penny.

I am really close to getting the answer and cant quite get it. I got C(w)= 60w^2+450/w for one and 60w^2+675/w for my other answer and it doesnt seem to be working on webwork. I got about 15 different answers from trying different stuff, but cant seem to find the right answer :(.
 
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A rectangular storage container with an open top is to have a volume of 30 cubic meters. The length of its base is twice the width. Material for the base costs 11 dollars per square meter. Material for the sides costs 5 dollars per square meter. Find the cost of materials for the cheapest such container. Answer needs to be rounded to the nearest penny.

I am really close to getting the answer and cant quite get it. I got C(w)= 60w^2+450/w for one and 60w^2+675/w for my other answer and it doesnt seem to be working on webwork. I got about 15 different answers from trying different stuff, but cant seem to find the right answer :(.



f(x)=x^(1/3)*(x+90)^(2/3)
This problem is a graphing problem where I need to find where it is increasing, decreasing, concave up and down, and inflection points. I have the critical numbers, but when I type it in on webwork, it doesnt seem to like my answers. Also need to be in interval notation.

IF anyone could help that would be greatly appreciated!!!!
Please only post one problem in a thread. You can edit this thread to delete the second question and start another thread with that question.

For the first question; if we let w be the width, b be the base (which is equal to 2w), and h the height, then the volume V is
V = w b h = 2 w2 h = 30
where w, b, and h are in meters. So
h = 15 / w2

The area of each of the front and back sides is b h = 2 w h = 30/w and the area of each of the left and right sides is w h = 15 / w. So the total area of the sides is 90 / w.

The area of the base is w b = 2 w2.

Therefore the cost function C is
C(w) = 22 w2 + 450 / w

Find the minimum, i.e. a point where C' = 0 and C'' > 0
 
Thanks you so much!!! This led me to the right answer! Now I know how to do it for the future. My teacher isnt really the best at explaining how to do problems :(
 
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