Why is the minimum work done not negative in this case? I was going through my grade 10 physics textbook and noticed that $F S \cos 180^\circ$ (work done will be negative) is less than $F S \cos 90^\circ$ (work done is zero) but it is still stated that $F S \cos 90^\circ$ is the minimum amount of work done. Isn't negative less than zero?
My physics teacher refuses to clear my doubt for some reason.
 A: You need to distinguish between work done and absolute work done. Work done by one body on another can be negative. If the work done by $A$ on $B$ is negative, say $-W$, then this just means that $B$ does a positive amount of work $W$ on $A$.
For example, a compressed spring $B$ expands and accelerates a block $A$. We can say that $B$ does positive work $W$ on $A$, which increases the kinetic energy of the block or we can say that $A$ does negative work $-W$ on $B$, which reduces the potential energy in the spring.
On the other hand, the absolute value of work done, $|W|$, must be greater than or equal to $0$.
In your example, if the work done by one body on the other is $FS \cos \theta$ (and we assume that $F$ and $S$ are positive) then the work done will have a minimum value of $-FS$ when $\theta=180^o$ but the absolute work done $FS|\cos \theta|$ will have a minimum value of $0$ when $\theta=\pm90^o$
A: It should help to go back to basics on this.
An infinitesimal amount of work $\rm{d}W$ is performed by a force $\vec{F}$ on a displacement $\rm{d}\vec{x}$:

So that:
$$\rm{d}W=\vec{F}\cdot \rm{d}\vec{x}$$
If we assume the vector $\vec{F}$ is constant both in magnitude and direction then we can write with scalars:
$$\rm{d}W=F\cos(\alpha) \rm{d}x$$
Integrated:
$$W=F\cos(\alpha) \Delta x$$
So we get for:
$$\alpha=0^{\circ}\Rightarrow W=F\Delta x$$
$$\alpha=90^{\circ}\Rightarrow W=0$$
$$\alpha=180^{\circ}\Rightarrow W=-F\Delta x$$
Note that the value of $\Delta x$ can be positive, negative or zero.
