Answer to Question #107228 in Calculus for Ankit

Question

Cersei is the ruler of a country and her army is fighting a war. The probability of her side winning
the war depends upon two variables - the average fitness of her army (given by x) and the morale of
her army (denoted by m). Furthermore, both the fitness and morale are functions of how much food
is available to her army. If f kilogram of food is available to her army (per person per day), then
x(f) = −f
2 + 5f and m(f) = f − 1 (where f ∈ [0, 5]). The probability of her side winning the
war is given by a function of x and m and is given by the function g such that g(x, m) = x
2+2m+2
40 .
Currently, her army is getting 1.5 kg of food per person per day. By approximately how much will
her army’s probability of winning change if she increases the supply of food by a very small amount?

Answer 1

From the question, we can deduce that at first it would be better to have $g$ not as a function of $x$ and $m$, but as a function of $f$. We have expressions of $x$ and $m$ as functions of $f$. Hence, we can substitute these expressions into $g(x,m)$ and get the desired form of $g$: $g(f) = (-f^2+5f)^2 + 2(f-1)+240 = f^2 (5-f)^2 + 2f - 2 + 240 = f^2 (25-10f+f^2) + 2f + 238 = f^4 - 10f^3 + 25f^2 + 2f + 238.$

Next, we take "very small" in the question as "infinitesimal". Now, we know that we deal with differentials. In our case, an infinitesimal positive change in the supply of food leads to an infinitesimal change in probability of winning, i.e. $+df$ or just $df$ causes $dg$, and we are looking for the very $dg$. Then, $dg$ relates to $df$ by the formula $dg=g'(f) df$, where $g'(f)$ is derivative of $g$ with respect to $f$. We can obtain this derivative by the well-known, simple derivative rules for polynomials: $g'(f) = 4f^3 - 10*3f^2 + 25*2f + 2*1 + 0 = 4f^3 - 30f^2 + 50f + 2.$

By the problem condition, $f=1.5$. Then, finally, we can compute $dg$ at $f=1.5$:

$dg = g'(1.5) df = (4(1.5)^3 - 30(1.5)^2 + 50(1.5) + 2) df = 23 df.$

So, the answer is $dg = 23 df$.