A common problem used to illustrate how differential calculus can be used for optimisation is "the box problem". The box problem goes as follows; imagine you manufacture boxes (W metres wide, D metres deep and H metres high) and you wish to minimise the amount of cardboard used to produce your standard box with volume V (V= W.D.H cubic metres).

The first step is to set up an area equation, which is the quantity that we are trying to minimize:

The first step is to set up an area equation, which is the quantity that we are trying to minimize:

A = (area of the two sides defined by the width) + (area of the two sides defined by the depth) + (area of the top and bottom sides)

=2W.H + 2D.H + 2W.D

Now we have three unknowns and two equations. One option is to form solution based on an assumed ratio (C) of the width to the height, which we can use to simplify our area equation to:

A = 2W.H. + 2(V/W) + 2(V/H) by using the the volume equation to substitute for D and using C= W/H, we get :

A= 2C.H^2 + 2 (V/C.H) + 2(V/H) = 2C.H^2 + (2/H)((V/C) + V)

A= 2C.H^2 + 2 (V/C.H) + 2(V/H) = 2C.H^2 + (2/H)((V/C) + V)

If we graph this function (A vrs H) and forget negative values of both A and H, we can see a clear vertical asymptote along the A = 0 and a minimum near the origin that is a function of our choices for V and C. Of course, this equation is ripe for differentiation:

dA/dH = 4C.H - (2/H^2)((V/C) + V)

At the minimum, it must follow:

dA/dH = 0 = 4.C.H - (2/H^2)((V/C) + V)

therefore,

H = ((V + VC)/(2 C^2))^1/3

Now, we have a ready way of optimising the quantity of cardboard for any given volume and ratio of height for depth. What solutions do we get if we assume a certain ratio to the width to the breadth ? Which is the true minimum (i.e. independent of our assumptions of ratios of dimensions) ? Excellent questions ! Start analysing and optimising ..... welcome to Applied Mathematics !

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