Start by creating a diagram.

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Create an expression:

1. Create an expression for perimeter \[\] \[x+x+y=3600\] \[2x+y=3600\]
2. Create an expression for area \[\] Since area of a rectangle is \(l \cdot w\), \(A\left(x\right)=xy\). \(A(x)\) is the function that represents the area. \[\]
3. Manipulate the area expression \[\] Before we do anything with the area function, we need to ensure that the area function is only dependent on one variable because, with our current knowledge, we can not do much with a function of two variables. Therefore, we need to solve the perimeter equation for a variable. Let's solve for \(y\). \[\mbox{Solve for y}\] \[y=3600-2x\] \[\mbox{Substitute that in for y in the area equation}\] \[A\left(x\right)=x\left(3600-2x\right)=3600x-2x^{2}\] We can now maximize the area equation. \[\]

Maximize the area expression:

1. Find the critical points \[\mbox{Find the derivative of \(A(x)\)}\] \[A'\left(x\right)=3600-4x\] \[\mbox{Set the function equal 0 and solve}\] \[3600=4x\] \[x=900\] Since the function is a polynomial (a function that is always continuous), the critical point we found is a guaranteed critical point. \[\]
2. Determine whether the critical point is either a maxima or a minima \[\] To do this we can either set up a sign diagram or use the second derivative. We will do both, but you can choose which one is more straightforward for you. \[\] \(\mbox{Sign Diagram}\) \[\]  Draw a sign diagram and determine what values we are going to test. \begin{array} {|r|r|}\hline x & x<900 &="" x="900">900 \\ \hline \mbox{sign of f'(x)} & f'(800) & 0 & f'(1000) \\ \hline  \end{array} \[\mbox{Complete the sign diagram}\] \begin{array} {|r|r|}\hline x & x<900 &="" x="900">900 \\ \hline \mbox{sign of f'(x)} & + & 0 & - \\ \hline  \end{array} Since the function changes from increasing to decreasing \(x=900\) is a maximum. \[\] \(\mbox{Second Derivative Test}\) \[\] \[\mbox{Find the second derivative}\] \[A''\left(x\right)=-4\] Since the second derivative is always negative, the original function is always concave down; therefore, \(x=900\) is a maximum. \[\] When asked to solve optimization problems you may be asked to show work, if so, use either of the methods shown (sign diagram or second derivative test). If you are not asked for work, you can determine if \(x=900 \) is a local maximum conceptually since the original function is a negative parabola. Negative or concave down parabolas only have a local maximum. </900></900>

Now that we know \(x\), we can find \(y\)

\[y=3600-2\left(900\right)=1800\]

Therefore, the dimensions of the field that will result in the greatest possible area are \(900 \cdot 1800\).

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*Your instructor may require you to add units to your answer. If so, the unit for this answer is \(m^2\), and your answer will be written as the following.

\[900\ m\cdot1800\ m=1620000\ m^{2}\]

Start by creating a diagram.

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Create an expression:

1. Create an expression for the length and width \[\] \[l=10-2x\] \[w=30-2x\] We are subtracting by \(2x\) because we are cutting \(x\) \(cm\) of material of of each side. \[h=x\] When the sides are folded up the height of the box will be the value of \(x\). \[\]
2. Create an expression for volume \[\] \[v=l \times w \times h\] \[\therefore\] \[v(x)=x\left(10-2x\right)\left(30-2x\right)\] \[v\left(x\right)=x\left(300-20x-60x+4x^{2}\right)\] \[=300x-80x^{2}+4x^{3}\]

Maximize the volume expression: 

1. Find the critical points \[\] \[\mbox{Find the derivative}\] \[v'\left(x\right)=12x^{2}-160x+300\] Since the function is a polynomial it will never be discontinuous, and all possible critical points we find will be critical points. \[\mbox{Using your calculator, find the where the derivative is equal to 0}\] \[12x^{2}-160x+300=0\] \[x=2.257\] \[x=11.076\] \[\]
2. Determine whether the critical points are either a maxima or a minima \[\] In the previous example we showed both methods, sign diagram and second derivative, for finding extrema. In this example we will only show the second derivative method because it is generally much quicker. If you would like to use the sign diagram method, you can refer back to the first example or to the extrema topic. \[\mbox{Find the second derivative}\] \[v''\left(x\right)=24x-160\] \[\mbox{Now find \(v''\left(2.257\right)\) and \(v''\left(11.076\right)\)}\] \[\mbox{using your calculator}\]\[v''\left(2.257\right)=-105.824\] \[v''\left(11.076\right)=105.824\]The local maximum occurs at \(v''\left(2.257\right)\) because at that point the function is concave down.

Since the second derivative at \(x=2.257\) is negative, the volume function at that point is concave down; therefore, \(x=2.257\) is a local maximum. At \(x=11.076\) the volume function is concave up; therefore, \(x=11.076\) is local minimum.

Therefore, when \(x=2.257\) \(cm\), the volume of the box will the greatest.

The dimensions of the box are \(10 \: cm \times 30 \: cm \times 2.257 \: cm\), and the volume of the box will be \(315.56\) \(cm^3\).

The formula for profit is:

\[p\left(x\right)=r\left(x\right)-c\left(x\right)\]

1. Find the function that models profit \[p\left(x\right)=60x-\left(x^{3}+\frac{5}{x}-12x\right)\] \[=60x-x^{3}-\frac{5}{x}+12x\] \[=72x-x^{3}-\frac{5}{x}\] \[=72x-x^{3}-5x^{-1}\]
2. Find the critical points \[\mbox{Find the derivative}\] \[p'\left(x\right)=72-3x^{2}+5x^{-2}\] The only point the derivative does not exist at is \(x=0\). Since the original function also does not exist at \(x=0\), \(x=0\) is not a critical point and we can ignore it. To find where the derivative is equal to \(0\), we will use our calculator. \[\mbox{Using your calculator, find the where the derivative is equal to 0}\] \[72-3x^{2}+5x^{-2}=0\] \[x=\pm 4.906\]
3. Determine whether the critical points are either a maxima or a minima \[\] We will use the second derivative test to identify the type of extrema. \[\mbox{Find the second derivative}\] \[p''\left(x\right)=-6x-10x^{-3}\] \[\mbox{Now find \(p''\left(4.906\right)\) and \(p''\left(-4.906\right)\)}\] \[\mbox{using your calculator}\] \[p''\left(4.906\right)=-29.5207\] \[p''\left(-4.906\right)=29.5207\]A trick for optimization questions is to ignore the negative critical values. For instance, in this example, \(x=-4.906\) is ignorable because it would mean that the company is producing negative products, which doesn't make sense; therefore, we only have to analyze \(x=4.906\). Since the second derivative at \(x=4.906\) is negative, the profit function at that point is concave down; therefore, \(x=4.906\) is a local maximum.
4. How many units need to be sold to maximize profits \[\]Since the \(x\) is in terms of thousands of units, we need to multiply \(4.906\) by a \(1000\) to determine the number of units that will result in the maximum profits. \[\mbox{Number of Units = 4906}\] \[\]

Now you can report to your boss that if the company sells \(\mbox{4,906}\) units, the company will maximize its profits.

Start by creating a diagram.

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In our diagram, we represented the length of the rectangle with \(2x\): however, using just \(x\) will also work to represent the whole length.

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Create an expression for the area of the rectangle:

\[l=2x\] \[w=y\] \[\therefore\] \[A(x)=2xy\]

Once again we can't maximize a function of two variables; therefore, we need to represent the area function in terms of only \(x\). To do this let's take a look at \(y\). Since the rectangle is inscribed inside the parabola, the width, \(y\), is just the value of the parabola at \(x\); therefore, \(y=16-x^2\). Let's substitute this in for \(y\) in the area function.

\[A(x)=2x\left(16-x^{2}\right)=32x-2x^{3}\]

Now we can maximize the area function.

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Maximize the area function:

1. Find the critical points \[\mbox{Find the derivative}\] \[A'\left(x\right)=32-6x^{2}\] Since the function is a polynomial it will never be discontinuous, and all possible critical points we find will be critical points.\[\mbox{Find where the derivative is equal to 0}\] \[32-6x^{2}=0\] \[-6x^{2}=-32\] \[x^{2}=\frac{32}{6}\] \[x=\pm\sqrt{\frac{32}{6}}\]
2. Determine whether the critical points are either a maxima or a minima \[\] We will do this using the second derivative test. \[\mbox{Find the second derivative}\] \[A''\left(x\right)=-12x\] \[\mbox{Find \(A''(\sqrt{\frac{32}{6}})\) and \(A''(-\sqrt{\frac{32}{6}})\)}\] \[A''\left(\sqrt{\frac{32}{6}}\right)=-12\sqrt{\frac{32}{6}}\] \[A''\left(-\sqrt{\frac{32}{6}}\right)=-12\left(-\sqrt{\frac{32}{6}}\right)=12\sqrt{\frac{32}{6}}\]Since \(-12\sqrt{\frac{32}{6}}\) is a negative number, \(A\left(\sqrt{\frac{32}{6}}\right)\) is concave down and \(A\left(\sqrt{\frac{32}{6}}\right)\) is a local maximum. Since \(12\sqrt{\frac{32}{6}}\) is a positive number, \(A\left(-\sqrt{\frac{32}{6}}\right)\) is concave down and \(A\left(-\sqrt{\frac{32}{6}}\right)\) is a local minimum. We can ignore \(x=-\sqrt{\frac{32}{6}}\) since we are only looking for the maximum area.

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The dimensions of the largest rectangle:

Since the maximum occurs when \(x=\sqrt{\frac{32}{6}}\), the value of \(x\) for the largest rectangle is also \(\sqrt{\frac{32}{6}}\).

\[\therefore\] \[\mbox{Since \(l=2x\)}\] \[l=2\sqrt{\frac{32}{6}}\]

\[\mbox{Since \(w=y=16-x^2\)}\] \[w=16-\left(\sqrt{\frac{32}{6}}\right)^{2}=16-\frac{32}{6}\] \[=\frac{96-32}{6}=\frac{64}{6}\]

The area of the largest rectangle is: \[A=2\sqrt{\frac{32}{6}}\left(\frac{64}{6}\right)\]

Start by creating a diagram.

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Once again, in our diagram, we represented the length of the rectangle with \(2x\): however, using just \(x\) will also work to represent the whole length.

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Find the equation of the positive semicircle:

The equation of a circle with a radius \(r\) is \(x^2+y^2=r^2\).

To get the equation we need to first solve for \(y\)

\[y^{2}=r^{2}-x^{2}\] \[y=\pm\sqrt{r^{2}-x^{2}}\]

The equation for a positive semicircle is just \(y=\sqrt{r^{2}-x^{2}}\)

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Create an expression for the area of the rectangle:

\[l=2x\] \[w=y\] \[\therefore\] \[A(x)=2xy\]

Once again we can't maximize a function of two variables; therefore, we need to represent the area function in terms of only \(x\). To do this let's take a look at \(y\). Since the rectangle is inscribed inside the parabola, the width, \(y\), is just the value of the semicircle at \(x\); therefore, \(y=\sqrt{r^{2}-x^{2}}\). Let's substitute this in for \(y\) in the area function.

\[A(x)=2x\sqrt{r^{2}-x^{2}}\]

Now we can maximize the area function.

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Maximize the area function:

1. Find the critical points \[\mbox{Find the derivative using the product rule}\] \[A'\left(x\right)=2\sqrt{r^{2}-x^{2}}+\left(2x\right)\frac{-2x}{2\sqrt{r^{2}-x^{2}}}\] \[=2\sqrt{r^{2}-x^{2}}+\frac{-2x^{2}}{\sqrt{r^{2}-x^{2}}}\] \[\mbox{Find where the derivative is equal to 0}\] \[2\sqrt{r^{2}-x^{2}}+\frac{-2x^{2}}{\sqrt{r^{2}-x^{2}}}=0\] \[\sqrt{r^{2}-x^{2}}\left(2\sqrt{r^{2}-x^{2}}+\frac{-2x^{2}}{\sqrt{r^{2}-x^{2}}}\right)=0\left(\sqrt{r^{2}-x^{2}}\right)\] \[2\left(r^{2}-x^{2}\right)-2x^{2}=0\] \[2r^{2}-4x^{2}=0\] \[2r^{2}=4x^{2}\] \[x^{2}=\frac{1}{2}r^{2}\] \[x=\pm\sqrt{\frac{1}{2}r^{2}}=\pm\sqrt{\frac{1}{2}}r\] \[\mbox{Rewrite the square root of the fraction using proper notation}\] \[x=\pm\sqrt{\frac{1}{2}}r=\pm\frac{\sqrt{1}}{\sqrt{2}}r=\pm\frac{1\sqrt{2}}{\sqrt{2}\sqrt{2}}r=\pm\frac{\sqrt{2}}{2}r\] The derivative also does not exist at two points where the original function exists. These are the endpoints of the semicircle. If we were to consider these point as possible value for the rectangle, the area of the rectangle would be \(0\) since the endpoints are on the \(x\) axis; therefore, we can ignore them. If you are interested, these point can be found by setting the denominator of the second term in the derivative to \(0\). \[\]
2. Determine whether the critical points are either a maxima or a minima \[\] We will do this using the second derivative test. Before we do that, let's make our lives easier and let \(u=r^{2}-x^{2}\) \[\mbox{Find the second derivative}\] \[A''\left(x\right)=\frac{2\left(-2x\right)}{2\sqrt{u}}+\frac{-4x\left(\sqrt{u}\right)-\left(-2x^{2}\right)\frac{u'}{2\sqrt{u}}}{\sqrt{u}^{2}}\] \[u'=-2x\] \[A''\left(x\right)=\frac{-4x}{2\sqrt{r^{2}-x^{2}}}+\frac{-4x\left(\sqrt{r^{2}-x^{2}}\right)-\left(-2x^{2}\right)\frac{-2x}{2\sqrt{r^{2}-x^{2}}}}{r^{2}-x^{2}}\] \[=\frac{-4x}{2\sqrt{r^{2}-x^{2}}}+\frac{-4x\sqrt{r^{2}-x^{2}}-\frac{2x^{3}}{\sqrt{r^{2}-x^{2}}}}{r^{2}-x^{2}}\] \[\mbox{Find \(A''(\frac{\sqrt{2}}{2}r)\) and \(A''(-\frac{\sqrt{2}}{2}r)\)}\] \[A''\left(\frac{\sqrt{2}}{2}r\right)=\frac{-4\frac{\sqrt{2}}{2}r}{2\sqrt{r^{2}-\frac{1}{2}r^{2}}}+\frac{-4\frac{\sqrt{2}}{2}r\sqrt{r^{2}-\frac{1}{2}r^{2}}-\frac{2\frac{\sqrt{2}}{2}r}{\sqrt{r^{2}-\frac{1}{2}r^{2}}}}{r^{2}-\frac{1}{2}r^{2}}\] Since the radius of a circle is is always greater than \(0\), let \(r=1\) to determine whether the above number is positive or negative. \[A''\left(\frac{\sqrt{2}}{2}\right)=\frac{-4\frac{\sqrt{2}}{2}}{2\sqrt{1^{2}-\frac{1}{2}1^{2}}}+\frac{-4\frac{\sqrt{2}}{2}\sqrt{1^{2}-\frac{1}{2}1^{2}}-\frac{2\frac{\sqrt{2}}{2}}{\sqrt{1^{2}-\frac{1}{2}1^{2}}}}{1^{2}-\frac{1}{2}1^{2}}\] \[=\frac{-2\sqrt{2}}{2\sqrt{\frac{1}{2}}}+\frac{-4\frac{\sqrt{2}}{2}\sqrt{\frac{1}{2}}-\frac{2\frac{\sqrt{2}}{2}}{\sqrt{\frac{1}{2}}}}{\frac{1}{2}}\] \[\mbox{Remember \(\sqrt{\frac{1}{2}}=\frac{\sqrt{2}}{2}\)}\] \[=\frac{-2\sqrt{2}}{2\frac{\sqrt{2}}{2}}+\frac{-4\frac{\sqrt{2}}{2}\frac{\sqrt{2}}{2}-\frac{2\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}}}{\frac{1}{2}}\] \[=-2+\frac{-2-2}{\frac{1}{2}}=-2+2\left(-4\right)=-10\] \[\] \[A''\left(-\frac{\sqrt{2}}{2}\right)=\frac{-4\left(-\frac{\sqrt{2}}{2}\right)}{2\sqrt{1^{2}-\frac{1}{2}1^{2}}}+\frac{-4\left(-\frac{\sqrt{2}}{2}\right)\sqrt{1^{2}-\frac{1}{2}1^{2}}-\frac{2\left(-\frac{\sqrt{2}}{2}\right)}{\sqrt{1^{2}-\frac{1}{2}1^{2}}}}{1^{2}-\frac{1}{2}1^{2}}\] \[=\frac{2\sqrt{2}}{2\sqrt{\frac{1}{2}}}+\frac{4\frac{\sqrt{2}}{2}\sqrt{\frac{1}{2}}-\frac{-2\frac{\sqrt{2}}{2}}{\sqrt{\frac{1}{2}}}}{\frac{1}{2}}\] \[=\frac{2\sqrt{2}}{2\frac{\sqrt{2}}{2}}+\frac{4\frac{\sqrt{2}}{2}\frac{\sqrt{2}}{2}+\frac{2\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}}}{\frac{1}{2}}\] \[=2+\frac{2+2}{\frac{1}{2}}=2+2\left(4\right)=10\] Since \(x=\frac{\sqrt{2}}{2}\) results in a negative number, \(x=\frac{\sqrt{2}}{2}r\) would've also result in a negative number. Therefore, \(x=\frac{\sqrt{2}}{2}r\) will be a local maximum. Since \(x=-\frac{\sqrt{2}}{2}\) results in a positive number, \(x=-\frac{\sqrt{2}}{2}r\) would've also result in a positive number. Therefore, \(x=-\frac{\sqrt{2}}{2}r\) will be a local minimum.

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Therefore, for the rectangle with the largest area, the value of \(x\) will be \(\frac{\sqrt{2}}{2}r\).

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The dimensions of the largest rectangle:

\[A\left(x\right)=2xy=2x\sqrt{r^{2}-x^{2}}\]

\[A\left(\frac{\sqrt{2}}{2}r\right)=\sqrt{2}r\sqrt{r^{2}-\frac{1}{2}r^{2}}=\sqrt{2}r\sqrt{\frac{1}{2}r^{2}}\] \[=\sqrt{2}r\frac{\sqrt{2}}{2}r=2r^{2}\]

Start by creating a diagram.

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We can do this problem one of two ways. We can either let the base be \(2x\) or we can realize that the area of a triangle is just half the area of a rectangle. Therefore, let's take one half of the triangle flip it and reflect it over the \(y\)-axis. The following is the result.

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As you can see, we are trying to find the largest rectangle that is both inscribed by the parabola and the first quadrant.

The area of the rectangle, \(A(x)\), is \(xy\).

Similar to the previous two examples, \(y\) is the value of the function at \(x\); therefore, \(y=25-x^{2}\). We can now substitute this into the area function and maximize it.

\[A\left(x\right)=x\left(25-x^{2}\right)=25x-x^{3}\]

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Maximize the Area function:

1. Find the critical points \[\mbox{Find the derivative}\] \[A'\left(x\right)=25-3x^{2}\]. Since the function is a polynomial it will never be discontinuous, and all possible critical points we find will be critical points. On top of that, since our rectangle is only in the first quadrant, we can ignore all critical points that lie on the negative axis.\[\mbox{Find where the derivative is equal to 0}\] \[25-3x^{2}=0\] \[25=3x^{2}\] \[x^{2}=\frac{25}{3}\] \[x=\pm\sqrt{\frac{25}{3}}=\pm\frac{5}{\sqrt{3}}\] \[=\pm\frac{5\sqrt{3}}{\sqrt{3}\sqrt{3}}=\pm\frac{5\sqrt{3}}{3}\]Since \(x=\frac{5\sqrt{3}}{3}\) is the only one that lies in first quadrant, it is the only critical point we need to analyze.
2. Determine whether the critical point is a maximum \[\] Since we are asked to maximize the area of the rectangle, it can be logically assumed that the only critical point is the maximum. However, since you may be asked to show work, we will prove this using the second derivative test. \[\mbox{Find the second derivative}\] \[A''\left(x\right)=-6x\] \[\mbox{Find \(A''\left(\frac{5\sqrt{3}}{3}\right)\)}\] \[A''\left(\frac{5\sqrt{3}}{3}\right)=-10\sqrt{3}\] Therefore, \(x=\frac{5\sqrt{3}}{3}\) is a local maximum since the function at that point is concave down.

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The dimensions of the largest rectangle:

The formula for the area of a triangle: \[A=\frac{1}{2}bh;\ b=2x,\ h=y=25-x^{2}\]

Using that, the dimension area of the largest triangle are:

\[b=2\left(\frac{5\sqrt{3}}{3}\right)\]

\[h=25-\left(\frac{5\sqrt{3}}{3}\right)^{2}=25-\frac{25\left(3\right)}{9}\]

\[=25-\frac{25}{3}=\frac{75}{3}-\frac{25}{3}=\frac{50}{3}\]

The area of the triangle is:

\[A\left(x\right)=\frac{1}{2}\left(2\left(\frac{5\sqrt{3}}{3}\right)\frac{50}{3}\right)=\frac{250\sqrt{3}}{9}\]

We are trying to minimize the surface area function; however, since we can't minimize a function of two variable, we need to represent either \(h\) or \(r\) in terms of the other variable. Let's represent \(h\) in terms of \(r\). To do this, we can use the volume formula.

\[v=\pi r^{2}h\]

\[\mbox{Now, solve for height}\]

\[h(r)=\frac{v}{\pi r^{2}}\]

\[\mbox{Substitute this in for r in the surface area equation}\]

\[SA(r)=2\pi r^{2}+2\pi r\left(\frac{v}{\pi r^{2}}\right)=2\pi r^{2}+2\left(\frac{v}{r}\right)\] \[=2\pi r^{2}+2vr^{-1}\]

Now, we can minimize the surface area function.

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Minimize the \(SA\) equation:

1. Find the critical points \[\mbox{Find the derivative}\] \[SA'\left(r\right)=4\pi r-2vr^{-2}=4\pi r-\frac{2}{r^{2}}v\] \[\mbox{Find where the derivative is equal to 0}\] From looking at the derivative, we can see that it does not exist at one one point, \(x=0\) (that's is where the second term does not exist). We can ignore \(x=0\) since the original function also does not exist there; therefore, all other critical points we find will be critical points. \[4\pi r-\frac{2}{r^{2}}v=0\] \[r^{2}\left(4\pi r-\frac{2}{r^{2}}v\right)=r^{2}\left(0\right)\] \[4\pi r^{3}-2v=0\] \[4\pi r^{3}=2v\] \[r^{3}=\frac{v}{2\pi}\] \[r=\sqrt[3]{\frac{v}{2\pi}}\]
2. Determine whether the critical point is a minimum \[\] Since we are asked to minimize the material used to produce the can, it can be logically assumed that the only critical point is the minimum. However, since you may be asked to show work, we will prove this using the second derivative test. \[\mbox{Find the second derivative}\] \[SA''\left(r\right)=4\pi+4vr^{-3}=4\pi+\frac{2}{r^{3}}v\] \[\mbox{Find \(\sqrt[3]{\frac{v}{2\pi}}\)}\] \[SA''\left(\sqrt[3]{\frac{v}{2\pi}}\right)=4\pi+\frac{2}{\left(\sqrt[3]{\frac{v}{2\pi}}\right)^{3}}v=4\pi+\frac{4\pi}{v}\]Since the volume is always positive, \(SA''\left(\sqrt[3]{\frac{v}{2\pi}}\right)\) is also always positive; therefore \(r=\sqrt[3]{\frac{v}{2\pi}}\) is a local minimum since the function at that point is concave up.

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The dimensions of the can:

\[r=\sqrt[3]{\frac{v}{2\pi}}\]

\[h(r)=\frac{v}{\pi r^{2}}\]

\[h\left(r\right)=\frac{v}{\pi\left(\sqrt[3]{\frac{v}{2\pi}}\right)^{2}}=\frac{v}{\pi\left(\frac{v}{2\pi}\right)^{\frac{2}{3}}}\]

Therefore, the surface area of a can with the lowest amount of material used possible is the following.

\[SA=2\pi r^{2}+2r\left(\frac{v}{\left(\frac{v}{2\pi}\right)^{\frac{2}{3}}}\right)\]