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Sagot :
(i) If R(m) = m² (c/2 - m/3) = cm²/2 - m³/3, then differentiating with respect to m gives the sensitivity function,
R'(m) = cm - m²
In Mathematica, define a function R by
R[m_] := m^2*(c/2 - m/3)
Then you can get the derivative with either R'[m] or D[R[m], m].
(ii) Find the critical points of R, i.e. the points for which R' is zero or undefined. Since R' is a polynomial, it is defined everywhere, so we only care about the first case. We have
R'(m) = cm - m^2 = m (c - m) = 0
with solutions m = 0 or m = c.
Compute the second derivative of R :
R''(m) = c - 2m
Check the sign of R'' at each critical point:
- R''(0) = c - 0 = c > 0, which indicates a minimum at m = 0
- R''(c) = c - 2c = -c < 0, which indicates a maximum at m = c
At the second critical point, we have a maximum value of
R(c) = c³/2 - c³/3 = c³/6
In M, we get the same result via
Maximize[R[m], m]
which should return a list like
{c^3/6, {m -> c}}
where the first element is the maximum, and the second element is a list with a rule pointing the target variable m to the critical point.
(iii) We've done this in part (ii) already while carrying out the second derivative test above,
R''(m) = c - 2m
In M, this is accomplished by R''[m] or D[R[m], {m, 2}].
(iv) We carry out the derivative test like in part (ii), but this time for R'. We have critical points when
R''(m) = c - 2m = 0
which occurs for m = c/2. The second derivative of the sensitivity function is the third derivative of R,
R'''(m) = -2
which is negative for all values of m, which means R' is in fact maximized at m = c/2. At this point of maximum sensitivity, we have
R(c/2) = c(c/2)²/2 - (c/2)³/3 = c³/8 - c³/24 = c³/12
In M, run
Maximize[R'[m], m]
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