The model is, to put it nicely, implausible. It assumes that the seeds have been hidden in sets of $100$, perhaps by a mathematically minded squirrel. It also assumes that searching for Type 1 and for Type 2 cannot be done simultaneously. And the answer is obvious, since the good stuff is easiest to find. But let us continue, but not without lodging a complaint at this parody of applied mathematics.
The expected caloric find if we spend the proportion $x_1$ searching for Type 1, and $x_2$ for Type 2, is
$$500x_1^3+300x_2^5.$$
Substituting $1-x_1$ for $x_2$, we arrive at the function
$$V(x_1)=500x_1^3+300(1-x_1)^5.$$
Differentiation yields
$$1500(x_1^2-(1-x_1)^4)=0.$$
So we want to solve $x_1^2=(1-x_1)^4$. Since everything is non-negative, we get equivalently $x_1=(1-x_1)^2$, or equivalently $x_1^2-3x_1+1=0$. This has the solution $x_1=\frac{3-\sqrt{5}}{2}$, and therefore $x_2=\frac{-1+\sqrt{5}}{2}$. Calculation shows that this is a horrible strategy. By using the second derivative, or by examining the first derivative, we see that in fact this gives an absolute minimum caloric intake.
So we need to look at the endpoints. It is clear that best strategy is to look for Type 1 only.
The model is, to put it nicely, implausible. It assumes that the seeds have been hidden in sets of $100$, perhaps by a mathematically minded squirrel. It also assumes that searching for Type 1 and for Type 2 cannot be done simultaneously. And the answer is obvious, since the good stuff is easiest to find. But let us continue, but not without lodging a complaint at this parody of applied mathematics.
The expected caloric find if we spend the proportion $x_1$ searching for Type 1, and $x_2$ for Type 2, is $$500x_1^3+300x_2^5.$$ Substituting $1-x_1$ for $x_2$, we arrive at the function $$V(x_1)=500x_1^3+300(1-x_1)^5.$$ Differentiation yields $$1500(x_1^2-(1-x_1)^4)=0.$$ So we want to solve $x_1^2=(1-x_1)^4$. Since everything is non-negative, we get equivalently $x_1=(1-x_1)^2$, or equivalently $x_1^2-3x_1+1=0$. This has the solution $x_1=\frac{3-\sqrt{5}}{2}$, and therefore $x_2=\frac{-1+\sqrt{5}}{2}$. Calculation shows that this is a horrible strategy. By using the second derivative, or by examining the first derivative, we see that in fact this gives an absolute minimum caloric intake.
So we need to look at the endpoints. It is clear that best strategy is to look for Type 1 only.