Can you please show the process too. This is pretty confusing to me. to the cons
ID: 2891067 • Letter: C
Question
Can you please show the process too. This is pretty confusing to me.
to the constraint. Activity 10.25. A cylindrical soda can holds about 355 cc of liquid. In this activity, we want to find the dimen- (a) What are the variables in this problem? What restriction(s), if any, are there on these (b) What quantity do we want to optimize in this problem? What equation describes the (c) Find and the values of your variables that satisfy Equation (10.16) in the context of (d) Determine the dimensions of the pop can that give the desired solution to this con- sions of such a can that will minimize the surface area. variables? constraint? this problem. strained optimization problem The method of Lagrange multipliers also works for functions of more than two variables.Explanation / Answer
The dependent variable is the amount of aluminum. Essentially, you must minimize the surface area of the cylinder. Write the primary equation: the surface area is the area of the two ends (each r²) plus the area of the side or lateral area
to minimize: A = 2r² + 2rh
The primary equation contains two independent variables, r and h. Can you relate them in some way? Yes, the problem constraint is that the volume equals 355 ml (or 355 cm³). This gives the secondary equation:
V = r²h = 355
Since h occurs once in the primary equation, and as a linear term, it will be easy to eliminate. Solve the secondary equation for h:
h = 355 / r²
Substitute in the primary equation:
A = 2r² + 2r·355/r² A = 2r² + 710/r
There are no obvious constraints on the feasible domain, except that r and h must both be positive. Since the feasible domain has no endpoints, you need not check whether they are minima.
To try to find maxima or minima, differentiate:
dA/dr = 4r 710/r²
d²A/dr² = 4 + 1420/r³
Find critical numbers where dA/dr = 0 or does not exist. The derivative does not exist at r = 0; however you can disregard that because r = 0 is outside the feasible domain.
4r 710/r² = 0 r = cuberoot(710/4) 3.84 cm
Is this a minimum, a maximum, or neither? Since d²A/dr² is positive for all positive r, the critical point r = 3.84 cm must be a minimum.
The problem asked for the dimensions of the can with lowest surface area, which means that you also need the height. To find it, substitute r = 3.84 in the secondary equation and get h 7.67 cm.
Answer: A cylindrical can with volume 355 ml will use the least aluminum if its radius is about 3.84 cm and its height is about 7.67 cm.
Check: V = r²h = (3.84²)(7.67) = 355.3 cm³, the same as the required volume give or take a little rounding difference.
Because a real soda can is not exactly cylindrical, you can’t expect perfect agreement with these figures. I measured about 3.2 cm radius and 12.7 cm height on a can of Barq’s root beer, including the top and bottom extensions.
By the way, you may have noticed that the radius is about ½ the height. In fact, you can prove that the cylinder of a given fixed volume with the lowest surface area will always have r = h/2.Instead of setting V = 355 keep V as a letter and treat it as a constant. You will have r = cuberoot(V/2) and substituting V = r²h gives r = h/2.