You are asked to build an open cylindrical can with volume 180.5 cubic inches. you will cut the bottom from a square of metal and form the curved side by bending a rectangular sheet of metal. . find the exact value of the radius and height of the can that will minimize the total amount of material required for the square and the rectangle.

In order to derive the function o f the height of the cylinder, you will use this formula

V=πr2h

h=d h=d=2r h=2r

V=πr22r

V(r,h)=180.5=πr2h⟹h=180.5πr2

⟹A(r)=r(r+h)=r2+180.5πr

Then you can derive the function, however there would be changes in the height and radius,considering that the volume is constant. There is also another way of solving it

The function should be

A(r)=2r(2r+h)=4r2+361πr

A′(r)=8r−361πr2

if gloria has "carry over" minutes, (that is, any minutes paid for but not used in a month can be applied to her payment for minutes in the next month,) her total usage for the three months is: total minutes = 320 + 243 + 489 = 1052 minutes number of 200 minute blocks = 1052/200 = 5 r52 (that is 5 with a remainder of 52) rounding up the number of 200 minute blocks gives her usage for the three months as 6 blocks. the cost for six blocks is 6 x $25 = $150 she also has 200 - 52 or 148 carry over minutes if gloria pays each month and unused minutes are lost, applying the rounding up procedure to each month gives: first month usage of 320 minutes rounds up to 2 blocks of 200 minutes for a payment of 2 x $25 = $50 second month usage of 243 minutes rounds up to 2 blocks of 200 minutes for a payment of 2 x $25 = $50 third month usage of 489 minutes rounds up to 3 blocks of 200 minutes for a payment of 3 x $25 = $75 total payments for the three months are $50 + $50 + $75 = $175 get the plan with carry over minutes!