5.6 Solving Optimization Problems Homework Answers

The point of 5.6 isn't the algebra—it's the modeling. In the real world, engineers aren't given neat formulas; they have to build the constraints themselves.

Radius = ( \sqrt[3]50 ) cm, Height = ( 100 / (50^2/3) ) cm.

Solving optimization problems is often the "final boss" of an introductory Calculus course. Whether you are working through a textbook or looking for , the goal is always the same: finding the most efficient way to use resources, whether that’s maximizing area, minimizing cost, or optimizing time. The Goal of Optimization 5.6 solving optimization problems homework answers

Volume constraint: $\pi r^2 h = 100\pi \implies r^2 h = 100 \implies h = \frac100r^2$. Cost function (let side cost = 1, top/bottom cost = 2): $C = 2(\textarea of top/bottom) + 1(\textarea of side)$ $C = 2(2\pi r^2) + 1(2\pi r h) = 4\pi r^2 + 2\pi r h$. Substitute $h$: $C(r) = 4\pi r^2 + 2\pi r (\frac100r^2) = 4\pi r^2 + \frac200\pir$. Derivative: $C'(r) = 8\pi r - \frac200\pir^2 = 0 \implies 8\pi r = \frac200\pir^2 \implies r^3 = 25$.

A cylindrical can is to hold ( 100\pi ) cm³ of liquid. Find radius and height that minimize surface area (closed top & bottom). The point of 5

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Graphing the constraints, we get:

2x + 3y ≤ 240 x + 2y ≤ 180 x ≥ 0 y ≥ 0