For instance, a few weeks ago you could have gotten this as a standard max/min homework problem: You would probably automatically find the derivative $A'(r)$ (which you could equivalently write as $\dfrac)$, then find the critical points, then determine whether each represents a maximum or a minimum for the function, and so forth.
It has trouble with word problems, but if you can write down a word problem in math notation it shouldn't be an issue.
I also tried it on a weird fraction from an AP algebra exam, which it kind of failed at, but then I swiped over and it was showing me this graph, which included the correct answer: I love this app, not just because it would've helped 8th grade Paul out of a jam, but because it's such a computery use of computers.
Hence, you already know how to do all of the following steps; the only new part to maximization problems is what we did in Stage I above. We want to minimize the function $$ A(r) = 2\pi r^2 \frac$$ and so of course we must take the derivative, and then find the critical points. \[ \begin A'(r) &= \dfrac\left(2\pi r^2 \frac \right) \\[8px] &= \dfrac\left(2\pi r^2 \right) \dfrac\left(\frac \right) \\[8px] &= 2\pi \dfrac\left(r^2 \right) 2V \dfrac\left(r^ \right) \\[8px] &= 2\pi(2r) 2V \left((-1)r^ \right)\\[8px] &= 4 \pi r\, – \frac \end \]The critical points occur when $A'(r) = 0$: \[ \begin A'(r) = 0 &= 4 \pi r\, – \frac \\[8px] \frac &= 4 \pi r \\[8px] \frac &= r^3 \\[8px] r^3 &= \frac \\[8px] r &= \sqrt \end \] We thus have only one critical point to examine, at $r = \sqrt\,.$Step 5.
Next we must justify that the critical point we’ve found represents a minimum for the can’s surface area (as opposed to a maximum, or a saddle point).
We’ve labeled the can’s height Having drawn the picture, the next step is to write an equation for the quantity we want to optimize.
Most frequently you’ll use your everyday knowledge of geometry for this step.Typical phrases that indicate an Optimization problem include: Before you can look for that max/min value, you first have to develop the function that you’re going to optimize.There are thus two distinct Stages to completely solve these problems—something most students don’t initially realize [Ref]. Now maximize or minimize the function you just developed.” We have provided those two dimensions, and so we are done.$\checkmark$We’ve now illustrated the steps we use to solve every single Optimization problem we encounter, and they always work. The first does not involve Calculus at all; the second is identical to what you did for max/min problems.We’ve already found the relevant radius, $r = \sqrt\,.$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 $r = \sqrt\,,$ \[ \begin h &= \frac\,\frac \[8px] &= \frac\,\frac \[8px] &= \frac\,\frac \[8px] &= 2^\frac \[8px] h &= 2^\sqrt \quad \triangleleft \end \] 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 \[8px] &= 2 \cdot\frac\,\sqrt \[8px] &= 2 \sqrt = 2r \end \] since recall that the ideal radius is $r = \sqrt\,.$ Hence the ideal height (height and radius) will minimize the cost of metal to construct the can?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). I was homeschooled (that's not the confession part), and in 8th grade my algebra textbook had the answers to half the problems in the back. That seems to be the premise behind app called Socratic. The app lets you take a picture of a problem (you can also type it in, but that's a little laborious), and it'll not only give you an answer, but the steps necessary to to arrive at that answer — and even detailed explanations of the steps and concepts if you need them.