*The first stage doesn’t involve Calculus at all, while by contrast the second stage is just a max/min problem that you recently learned how to solve: single variable. You’ll use your usual Calculus tools to find the critical points, determine whether each is a maximum or minimum, and so forth. In Optimization problems, always begin by sketching the situation. If nothing else, this step means you’re not staring at a blank piece of paper; instead you’ve started to craft your solution.The problem asks us to minimize the cost of the metal used to construct the can, so we’ve shown each piece of metal separately: the can’s circular top, cylindrical side, and circular bottom.You'll see a button "View steps" and this takes you to the developer's site where you can purchase the full version of the solver (where you can see the steps).*

Above, for instance, our equation for $A_\text$ has two variables, We can now make this substitution $h = \dfrac$ into the equation we developed earlier for the can’s total area: \[ \begin A_\text &= 2\pi r^2 2 \pi r h \\[8px] &= 2\pi r^2 2 \pi r \left( \frac\right) \\[8px] &= 2\pi r^2 2 \cancel \cancel \left(\frac\right) \\[8px] &= 2\pi r^2 \frac \end \]We’re done with Step 3: we now have the function in terms of a single variable, , and we’ve dropped the subscript “total” from $A_\text$ since we no longer need it.

This also concludes Stage I of our work: in these threes steps, we’ve developed the function we’re now going to minimize!

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We’ve already found the relevant radius, $r = \sqrt[3]\,.$To find the corresponding height, recall that in the Subproblem above we found that since the can must hold a volume of liquid, its height is related to its radius according to $$h = \dfrac\,.

$$ Hence when $r = \sqrt[3]\,,$ \[ \begin h &= \frac\,\frac \[8px] &= \frac\,\frac \[8px] &= \frac\,\frac \[8px] &= 2^\frac \[8px] h &= 2^\sqrt[3] \quad \triangleleft \end \] The preceding expression for is correct, but we can gain a nice insight by noticing that $^ = 2 \cdot\frac$$ and so \[ \begin h &= 2^\sqrt[3] \[8px] &= 2 \cdot\frac\,\sqrt[3] \[8px] &= 2 \sqrt[3] = 2r \end \] since recall that the ideal radius is $r = \sqrt[3]\,.$ Hence the ideal height (height and radius) will minimize the cost of metal to construct the can?

Notice, by the way, that so far in our solution we haven’t used any Calculus at all.

That will always be the case when you solve an Optimization problem: you don’t use Calculus until you come to Stage II.

first have to develop the function you’re going to maximize or minimize, as we did in Stage I above.

Having done that, the remaining steps are exactly the same as they are for the max/min problems you recently learned how to solve.

## Comments How To Solve Calculus Word Problems

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