Due to the increasing cost of aluminum a brewery wants to redesign their cylindrical tanks
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I'm having trouble solving this simple optimization problem, can't work out where I'm going wrong.
A brewery wants to make a cylindrical aluminium beer can which will hold 375ml. (This means the volume of the can is 375 cm3 ). (Assume that any aluminium used for the joins and tab are not included in the calculations.) Set up an appropriate mathematical model which can be used to calculate the radius of the base of the can if the amount of aluminium used in its construction is to be minimised.
So I think: write an equation for the volume (V) and area (A) based on the radius & height:
375=V=πhr2375=V=πhr2
A=2πr2+πhr2A=2πr2+πhr2
Let's use the constant V to simply to a single variable:
h=Vπr2h=Vπr2
A=2πr2+πr2Vπr2A=2πr2+πr2Vπr2
A=2πr2+VA=2πr2+V
To find the optimal value of r which minimizes A, we take the first derivative and solve for 0:
dAdr=4πrdAdr=4πr
4πr=04πr=0
Therefore the optimal value for r to minimize A is:
A brewery wants to make a cylindrical aluminium beer can which will hold 375ml. (This means the volume of the can is 375 cm3 ). (Assume that any aluminium used for the joins and tab are not included in the calculations.) Set up an appropriate mathematical model which can be used to calculate the radius of the base of the can if the amount of aluminium used in its construction is to be minimised.
So I think: write an equation for the volume (V) and area (A) based on the radius & height:
375=V=πhr2375=V=πhr2
A=2πr2+πhr2A=2πr2+πhr2
Let's use the constant V to simply to a single variable:
h=Vπr2h=Vπr2
A=2πr2+πr2Vπr2A=2πr2+πr2Vπr2
A=2πr2+VA=2πr2+V
To find the optimal value of r which minimizes A, we take the first derivative and solve for 0:
dAdr=4πrdAdr=4πr
4πr=04πr=0
Therefore the optimal value for r to minimize A is:
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