Probem14+Elana,+Zach+F.



The problem reads as follows: A rectangular storage container with an open top is to have a volume of 10m cubed. The length of its base is twice the width. Material for the base costs $10 per square meter. Material for the sides costs $6 per square meter. Find the cost of materials for the cheapest such container.

The first equation we need to write down is that of the volume of a rectangular container, V = l*w*h. Since we know that the Volume of the equation is 10 m cubed, and we also know that the length of the base is twice the width. we can set the equation in terms of width and height alone. and we get the equation 10 = 2w*w*h. thus 10 = 2w^2*h and so. h =5/(2w^2) and l = 2w check this -- I get h = 5/(w² )

we are trying to minimize the cost of the storage container. We are going to set up the following equation C = l*w*10 + w*h*2*6 + l*h*2*6 + l*w*6 C = 20w^2 + 60/w + 120/w + 12w^2 C = 32w^2 + 180/w

then you have to take the derivative to find the critical values C' = 64w - 180/w^2 = 0 64w = 180/w^2 w = 1.412 meters

then plug that back into C

C = 2(1.412^2)10 + ... = $191.29 and that is your answer.

Not what I get. Look at how you are representing payment for the front and back (two sides there) and left and right (again, two sides).