A factory produces `x` units of model P and `y` units of model Q. Each model P requires `a` hours of assembly and `b` u…

Let `x` be the number of units of model P and `y` be the number of units of model Q. The problem states several constraints:

1. Non-negativity: `x ≥ 0` and `y ≥ 0`

2. Component usage: The total components used must be exactly `C`. This gives the equation `bx + dy = C`.

3. Assembly time: The total assembly hours must not exceed `H`. This gives the inequality `ax + cy ≤ H`.

Our goal is to find the condition on `H` and `C` that guarantees there exists at least one pair `(x, y)` satisfying all these constraints.

From the component equation, we can express `y` in terms of `x`: `y = (C - bx) / d`.

Since `y` must be non-negative, `(C - bx) / d ≥ 0`. As `d` is positive, this implies `C - bx ≥ 0`, or `x ≤ C/b`. Combined with `x ≥ 0`, the possible values for `x` are in the interval `[0, C/b]`. A possible plan exists if we can find an `x` in this interval that also satisfies the assembly time inequality.

Let's define the assembly hours needed as a function of `x`, `A(x)`. Substitute the expression for `y` into the assembly inequality:

`A(x) = ax + c((C - bx)/d) ≤ H`

This function `A(x)` is a linear function of `x`. A linear function's value over a closed interval is always between its values at the endpoints.

So, we need to find the range of `A(x)` for `x` in `[0, C/b]`.

Calculate `A(x)` at the endpoints:

- At `x = 0` (producing only model Q): `A(0) = a(0) + c(C/d) = cC/d`.

- At `x = C/b` (producing only model P): `A(C/b) = a(C/b) + c((C - b(C/b))/d) = a(C/b) + c(0) = aC/b`.

All possible values for the required assembly hours, `A(x)`, lie in the interval between `aC/b` and `cC/d`. For a plan to be possible, at least one of these required assembly hour values must be less than or equal to the available hours, `H`. This is only possible if the · minimum · of the possible required hours is no more than `H`.

The minimum value in the range is `min(aC/b, cC/d)`.

Therefore, the condition for a plan to be possible is:

`min(aC/b, cC/d) ≤ H`

Since `C` is a positive constant, we can divide both sides by `C` without changing the inequality direction:

`min(a/b, c/d) ≤ H/C`

This is equivalent to `H/C ≥ min(a/b, c/d)`.

Choice A is incorrect because requiring `H/C` to be greater than or equal to the `max` would mean that · every · possible production plan that uses all `C` components is viable, but the question only asks for the existence of · at least one · such plan.

Choice C is incorrect because it reverses the inequality, which would likely result from an algebraic error.

Choice D is incorrect because it uses the ratio `(a+c)/(b+d)`, a common fallacy that attempts to find an 'average' or 'combined' ratio without analyzing the system of constraints correctly; this value does not represent a boundary of the feasible set of solutions.