A factory produces two types of components, A and B. The cost to produce x units of component A is C_A(x) = 2x^2 + 500, and the cost to produce y units of component B is C_B(y) = 3y^2 + 200. If the factory produces a total of 100 units (x + y = 100) and the total cost is minimized, which equation must be true for the number of units produced?
- Ax - y = 0
- B2x + 3y = 0
- C4x = 6y
- Dx + y = 100
Show answer & explanationAnswer & explanation
Correct answer: C. 4x = 6y
This is an optimization problem involving two variables subject to a constraint. We need to minimize total cost C(x, y) = C_A(x) + C_B(y) = 2x^2 + 500 + 3y^2 + 200. Constraint: x + y = 100, so y = 100 - x. Substitute y into the cost function: C(x) = 2x^2 + 3(100 - x)^2 + 700. To minimize, find the derivative and set to zero: C'(x) = 4x + 6(100 - x)(-1) = 4x - 600 + 6x = 10x - 600. Set 10x - 600 = 0, so 10x = 600, x = 60. If x = 60, then y = 40. Now check the options. Option A: 4x = 6y => 4(60) = 6(40) => 240 = 240. This is true. (Alternatively, for minimum cost, marginal costs must be equal: C_A'(x) = 4x, C_B'(y) = 6y, so 4x = 6y).
Why the other options are wrong
- A. This implies x=y, which would mean 50 units of each, but this does not minimize the cost given the different quadratic coefficients.
- B. This equation is incorrect and does not reflect an optimization condition.
- D. This is the constraint, not the condition for minimized cost.
Optimization with Constraints
Optimization problems involve finding the maximum or minimum value of a function, often subject to one or more constraints that limit the possible values of the variables.
- Can often be solved by substituting the constraint into the objective function.
- Calculus (derivatives) is used to find critical points (max/min).
- For two variables, marginal rates (derivatives with respect to each variable) are often equal at the optimum.
Memory trick: Link the 'Constraint' to the 'Objective', then 'Derive' to find the 'Extrema'.