Introduction to Optimization Problems AB & BC
An optimization problem carries an objective, the quantity to be made largest or smallest, and a constraint, the relationship that limits the choices. The question decides which is which. For a rectangular plot with 1200 feet of fence on three sides, the objective is $A = xy$ and the constraint is $2x + y = 1200$; solving the constraint for $y$ and substituting gives $A(x) = x(1200 - 2x)$, a single function of a single variable that the rest of the unit can handle.
Two habits break the setup. The constraint goes unused, and $A = xy$ is differentiated with $y$ treated as a constant, which asserts that lengthening one side leaves the other alone. And the domain is taken from the algebra rather than from the situation: $V(x) = x(12 - 2x)^{2}$ is a polynomial defined everywhere and describes a real box only for $0 < x < 6$. Positivity, stated caps, whole-number counts, and the point where a formula stops applying are all part of the domain. When the objective is a distance, minimize its square, since both are smallest at the same input.
The work
3 ways in · any order
Lesson
Introduction to Optimization Problems
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Separates the objective from the constraint and shows why the question rather than the algebra decides which is which, reduces two-variable objectives to one variable on the fence, box, and can problems, derives each domain from the physical situation, and adds the move that makes distance problems tractable.
Diagnostic
10-item topic check
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Ten items spanning the two failure modes of this topic: differentiating an objective in two variables without using the constraint to eliminate one, and optimizing over all real numbers while ignoring the restrictions and endpoints the context imposes.
Targeted Practice
Drill a single misconception
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Pick one of the failure modes you missed and drill it on its own. The round is adaptive: two correct in a row clears the misconception and moves you to the next.