*Let’s solve some common problems step-by-step so you can learn to solve them routinely for yourself.*

Recall that $\dfrac\left(x^n\right) = nx^.$ \[ \begin \dfrac\left( \fracx^9\right) &= \frac \dfrac\left(x^9 \right) \\[8px] &= \frac\left(9 x^ \right) \\[8px] &= \frac(9) \left(x^8 \right) \\[8px] &= 6x^8 \quad \cmark \end \] Recall that $\dfrac\left(x^n\right) = nx^.$ We simply go term by term: \[ \begin \dfrac\left(2x^3 – 4x^2 x -33 \right) &= \dfrac\left( 2x^3\right) – \dfrac\left( 4x^2\right) \dfrac(x) – \cancelto \\[8px] &= 2\dfrac\left( x^3\right) – 4\dfrac\left( x^2\right) \dfrac(x) \\[8px] &= 2 \left(3 x^2 \right) – 4 \left(2x^1 \right) 1 \\[8px] &= 6x^2 -8x 1 \quad \cmark \end \] Recall that $\dfrac\left(x^n\right) = nx^.$ The rule also holds for fractional powers: \[ \begin \dfrac\left(\sqrt \right) &= \dfrac\left(x^ \right) \\[8px] &= \fracx^ \\[8px] &= \fracx^ \quad \cmark \\[8px] &= \frac\frac \quad \cmark \end \] Note that the last two lines are completely equivalent, and either would be acceptable as the answer.

Recall that $\dfrac\left(x^n\right) = nx^.$ The rule also holds for negative powers: \[ \begin \dfrac \left(\dfrac \right) &= 5 \dfrac \left(x^ \right) \\[8px] &= 5 \left((-3) x^ \right) \\[8px] &= -15 x^ \quad \cmark \\[8px] &= \frac \quad \cmark \end \] Note that the last two lines are completely equivalent, and either would be acceptable as the answer.

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.

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 $A'(r) = 0$: \[ \begin A'(r) = 0 &= 4 \pi r\, – \frac \\[8px] \frac &= 4 \pi r \\[8px] \frac &= r^3 \\[8px] r^3 &= \frac \\[8px] r &= \sqrt[3] \end \] We thus have only one critical point to examine, at $r = \sqrt[3]\,.$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 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.

Constants come out in front of the derivative, unaffected: $$\dfrac\left[c f(x) \right] = c \dfracf(x) $$ For example, $\dfrac\left(4x^3\right) = 4 \dfrac\left(x^3 \right) =\, …

$The derivative of a sum is the sum of the derivatives: $$\dfrac \left[f(x) g(x) \right] = \dfracf(x) \dfracg(x) $$For example, $\dfrac\left(x^2 \cos x \right) = \dfrac\left( x^2\right) \dfrac(\cos x) = \, …$\begin \dfrac(fg)&= \left(\dfracf \right)g f\left(\dfracg \right)\[8px] &= \Big[\text \times \text\Big] \Big[\text \times \text\Big] \end \begin \dfrac\left(\dfrac \right) &= \dfrac \[8px] &=\dfrac \end Many students remember the quotient rule by thinking of the numerator as “hi,” the demoninator as “lo,” the derivative as “d,” and then singing “lo d-hi minus hi d-lo over lo-lo” We’ll show more detailed steps here than normal, since this is the first time we’re using the Power Rule.

## Comments Calculus Problems Solved Step By Step