Question
Starkist tuna company will package tuna fish in 2-L cans. Find the radius and the height of the cans if the cans have a surface area which is less that 980 cm^2.
I have no idea how to go about this, it seems like geometry to me. Can you give me some help? Please.
I have no idea how to go about this, it seems like geometry to me. Can you give me some help? Please.
Answers
bobpursley
Sarea=PI*diameter*height + 2PI*(diameter/2)^2
Volume=PI(dia/2)^2 h
put numbers in for volume, Sarea, then solve for height in the volume equation, put that in for height in the Area equation. Solve for diameter.
Volume=PI(dia/2)^2 h
put numbers in for volume, Sarea, then solve for height in the volume equation, put that in for height in the Area equation. Solve for diameter.
volume of can = pi(r^2)(h)
2000 = pi(r^2)(h)
Surface area of the can = 2 circles + 1 rectangle, where the length of the rectangle is the circumference of the circle and its height is the height of the can.
so
2pi(r^2) + 2rpi < 980
This is actually a typical Calculus problem, where the question would ask for the dimensions of the can with the minimum surface area to hold 2 L
The way it was worded, there would be an infinite number of solutions.
2000 = pi(r^2)(h)
Surface area of the can = 2 circles + 1 rectangle, where the length of the rectangle is the circumference of the circle and its height is the height of the can.
so
2pi(r^2) + 2rpi < 980
This is actually a typical Calculus problem, where the question would ask for the dimensions of the can with the minimum surface area to hold 2 L
The way it was worded, there would be an infinite number of solutions.
Is the volume 2L? How can I solve for diameter if I don't have the height?
How do I get the volume?
bobpursley
You solve for diameter after putting in for height (area/2pi(dia/2)^2 into the Area equation.
It is called substitution, having two equations and solving it by substitution.
It is called substitution, having two equations and solving it by substitution.
bobpursley
Zach. The volume was given as 2dm^3 (2liters). Use the 2dm^3, that is essential.
If you want the surface area to be equal to 980 and the volume equal to 2 L or 2000 cm^3
then you end up solving a nasty cubic
pir^3 - 980r + 4000 = 0
I found this handy cubic equation solver
http://www.1728.com/cubic.htm
and obtained two possible solutions:
r = 15.08 or r = 4.34
both give us a volume of 2 L and a surface area of 980 cm^2
So if you want your surface area to be < 980, then any r between 4.34 and 15.08 and its corresponding h = 2000/(pir^2) would give a volume of 2 L and a surface area less than 980 cm^2
then you end up solving a nasty cubic
pir^3 - 980r + 4000 = 0
I found this handy cubic equation solver
http://www.1728.com/cubic.htm
and obtained two possible solutions:
r = 15.08 or r = 4.34
both give us a volume of 2 L and a surface area of 980 cm^2
So if you want your surface area to be < 980, then any r between 4.34 and 15.08 and its corresponding h = 2000/(pir^2) would give a volume of 2 L and a surface area less than 980 cm^2
I tried to solve for H in the volume equation. so, 2=3.14(r^2)H
divide 2 by 3.14=.64
.64=r^2H
.64/r^2=H
Not sure where to go from here
divide 2 by 3.14=.64
.64=r^2H
.64/r^2=H
Not sure where to go from here
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