Optimization exercise for advanced placement calculus classes (AP Calculus) : minimize the surface area of a cylindrical tank made of steel (with a given volume) in order to get the dimensions that save a maximum of raw material. Mathematical tools to use: first derivative and second derivative, methods for solving a system of linear equations, formula to compute a disc area and formula for finding the volume of a cylinder.
Mathematics and derivatives : Solution exercise 1.5 (Advanced placement math level)
1.5 Optimization : Henry Ford and its moving assembly line
In a factory where cars are being manufactured on an Assembly Line, the car is assembled thanks to the genius of the American founder of the Ford Motor Company (Henry Ford was openly sympathetic to the Nazis and Anti-Semite,... well, a bad guy. Nevertheless he invented and created the first moving assembly line in 1906). However, before the car is assembled, the doors of the car must be painted. Therefore a huge steel cylindrical tank contains 64 m3 of paint (when it is full). You are asked to find the dimensions of that cylindrical tank in order to minimize the quantity of steel required to make it. The point is to save raw material and thus Earth's limited resources. Important note: the cylindrical tank is closed at both ends by a lid.
Let us draw a schema of the cylindrical tank
To find the surface area of the cylinder you need to add the surface area of both end (the lids) and the surface area of the side. The surface area of the cylinder is symbolized by
A(r), which is a function of radius r:
A(r) = 2 π r2 + 2 π r h
Summary of the equations we have found:
(1) Volume of the cylinder: Vol = π r2 h = 64 m3
(2) Surface area : A(r) = Lid area + Side area = 2 π r2 + 2 π r h
Isolate h in equation (1): h = 64 / (π r2)
Substitute 64 / (π r2) for h in equation (2): A(r) = 2 π r2 + 2 π r ( 64 / (π r2) )
A(r) = 2 π r2 + 128 / r
We want to get the cheapest possible tank with a volume of 64 m3. In other words we must use the smallest possible quantity of metal. To do so we will have to minimize the surface area of the tank and to calculate the radius r and height h that give a A(r)min (minimum surface area).
Reminder:
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How to find the x and y coordinates of an extrema ? Take the first derivative of the function set equal to zero and solve for x. In fact the x-value that makes the first derivative = zero is the x-coordinate of an extrema (minimum or maximum). If the second derivative of a function is positive, then the function is concave up, which means that the curve is curving upwards, like a smile (remember: when you are up, you smile). Hence a minimum implies a positive second derivative. While a negative second derivative implies a concave down curve, which means that the curve is curving downwards like a frowning face (remember: when you are down, you frown). Hence a maximum implies a negative second derivative.
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We must then solve the following system of equations for r:
A'(r) = 0 and
A' '(r) > 0
A(r) = 2 π r2 + 128 / r
We can derivate easily this mathematical expression with the 18 derivative rules and we get:
A '(r) = 4 π r - 128 / r2 = 0
thus 4 π r3 = 128
r3 = 128 / 4 π
The second derivative is A' '(r) = 4 π + 256 / r3. Therefore, no matter the value of the radius r, the second derivative A' '(r) will always be positive and we will always have a minimum.