7. Optimization and Approximation Free Online FlashCards

What is optimization?

Optimization finds the largest or smallest possible value of an objective quantity subject to one or more constraints.

What is the general strategy for an optimization problem?

The main steps are to identify the objective, define variables, write constraints, form a one-variable objective function, determine its domain, find critical points, test endpoints, and interpret the result.

What rectangle maximizes area with perimeter 40 m?

For a rectangle with perimeter 40 m, the maximum-area dimensions are 10 m by 10 m, giving an area of 100 m².

What does the Extreme Value Theorem guarantee on [a,b]?

If f is continuous on [a,b], it has an absolute maximum and minimum there; each occurs at an endpoint or at an interior critical point.

How do you find absolute extrema on a closed interval?

Evaluate the function at both endpoints and every critical point in the interval, then compare all values. The largest is the absolute maximum and the smallest is the absolute minimum.

How does fixed volume constrain a cylinder’s dimensions?

For a cylinder with fixed volume V, πr²h = V, so h = V/(πr²). This constraint lets surface area be written as a function of r alone.

How are revenue and profit modeled in terms of quantity?

Revenue is R(q) = q p(q), and profit is P(q) = R(q) − C(q), where q is quantity, p(q) is price, and C(q) is cost.

When does the projectile h(t) = −16t² + 64t + 5 reach maximum height?

For h(t) = −16t² + 64t + 5, the maximum occurs at t = 2 because h′(t) = −32t + 64 is zero; the maximum height is 69 ft.

What is the linearization of f at x = a?

The linearization of f at x = a is L(x) = f(a) + f′(a)(x − a). Near a, f(x) is approximately L(x).

How does linearization approximate √4.1?

Using f(x) = √x at a = 4 gives L(x) = 2 + ¼(x − 4), so √4.1 ≈ L(4.1) = 2.025.

What is the differential of y = f(x)?

For y = f(x), the differential is dy = f′(x) dx. When dx is small, dy approximates the actual change Δy.

How does a small radius change affect circle area?

For A = πr², dA = 2πr dr. At r = 10 cm and dr = 0.02 cm, dA = 0.4π cm² ≈ 1.257 cm².