Critical Points in 2D

Business Calculus · Axiom Academy

EXAMPLE Finding Critical Points in 2D Worked examples of solving systems of partial derivatives Step 1: Find the partial derivatives Step 2: Set both equal to zero Step 3: Solve using substitution Case 1: x = 0. Then from (2): -2y = 0, so y = 0 → Point (0, 0) Case 2: x ≠ 0, so 3x - 6 = 0, giving x = 2. Then from (2): 2(2) - 2y = 0, so y = 2 → Point (2, 2) When you get factored expressions, consider all cases where each factor could equal zero! A company's profit from two products is: where x and y are quantities produced (in hundreds) Step 2: Set marginal profits to zero Step 3: Solve the linear system This is a 2×2 linear system. Multiply equation (1) by 2: Add to equation (2) to eliminate y: Produce 1,800 units of product 1 and 1,000 units of product 2 Step 1: Find the partial derivatives (using product rule) Since is never zero, we can divide it out: Since e^(anything) is always positive, it can never equal zero. This simplifies solving by allowing us to divide by the exponential factor. Based on the example above, which approach is correct?

This is the written version of the interactive lesson above. See the full Business Calculus course.