To maximise lifetime reproductive success, a tree must optimise the allocation of its resources between growth and defence against predators. A model of resource allocation requires the maximisation of the functional 00 S[Y] = = -5* 0 where: dx p(x) M(Y(x)), • • x is the age of the tree; y(x) = [0, 1] is the proportion of resources given to growth at age x (so that 1 − y(x) is the proportion given to defence); - Y(x)= dt y(t) is the cumulative proportion given to growth (so that Y'(x) = y(x) and Y(0) = 0); M(Y(x)) is the rate of reproduction, which satisfies M(0) = 0; p(x) is the probability of a tree remaining alive at age x- note that in this model p(0) = 1, 0 < p(x) < 1 for 0 < x < ∞, p(∞) = 0, and p satisfies the differential equation p'(x) = − (µ(x) − y(1 − y(x))) p(x), where μ(x) > 0, the death rate at age x, is a known function, and 20 is a constant. Throughout this question you may assume that M and p are differentiable to any order required and that the usual methods of the Calculus of Variations hold for functions defined on the infinite interval [0, ∞). (a) Show that p(x) = exp (-5* 'dt µ(t) + v(x − Y(x))). 0 ((b) Show that the Euler-Lagrange equation for S[Y] in the range x > 0 is Can show you more details about the proof M'(Y) M(Y) = 0. Ans (b) Let [S][Y] = √ √ ³° dx F(x,Y), where F(Y) = p(x) M(Y(x)). The Euler-Lagrange equation is OF ΟΥ = др 0, i.e. M(Y)+p M'(Y) = 0. Now, Op/OY: ΟΥ = -yp, so M'(Y) M(Y) = 0, as required, since p(x) > 0.