Optimization Problems - Calculus

We start by writing the integral in terms of \(t\).

We follow our four-step method:

• Step 1: The quantity we're trying to optimize (maximize in this case) is the rectangular surface area of Paul's enclosure. The variables are the width \(x\) and the "height" \(y\) of the rectangle making the enclosure. The enclosed area is therefore: \[A = xy\]
• Step 2: The constraint is the fact that Paul only has 40m of fencing. Those 40m are linked to the variables \(x\) and \(y\) by the fact that: \[40 = 2x+y\]
• Step 3: rearranging \(40 = 2x + y\) we can make \(y\) the subject: \[y = 40 - 2x\] Now we can replace \(y\) in our expression for the area, \(A=xy\), to express \(A\) in terms of \(x\) only: \[A = x\begin{pmatrix}40 - 2x\end{pmatrix}\] that's: \[A = 40x - 2x^2\] which we can write as a function of \(x\): \[A(x) = 40x - 2x^2\]
• Step 4: to find the maximum area, we turn to calculus and start by differentiating \(A(x)\): \[A'(x)=40 - 4x\] We now solve \(A'(x)=0\), that's: \[\begin{aligned} & 40 - 4x = 0 \\ & 40 = 4x \\ & 10 = x \\ & x = 10 \end{aligned}\] To check that \(x=10\) corresponds to a maximum, we can use the second derivative test: \[A''(x) = -4\] So when \(x = 10\): \[A''(10) = -4 < 0 \] Since the second derivative is negative, we can confirm: \(x=10\) corresponds to the maximum. Using the expression found at the beginning of step 3 we can therefore state that: \[\begin{aligned} y & = 40 - 2\times 10 \\ & = 40 - 20 \\ y & = 20 \end{aligned}\]

• width : \(x = 10\)
• "height" : \(y= 20\)

We follow our four-step method:

• Step 1: The quantity we're trying to optimize (minimize in this case) is the soda can's surface area, which we can call \(S\). The soda can is modelled by a cylinder whose variables are its radius \(r\) and its height \(h\). The can's surface area, \(S\), consists of:
  • the sum of two discs, one at the top, the other at the bottom: \(\pi r^2 + \pi r^2 = 2\pi r^2\)
  • the curved surface area of the can: \(2\pi r h\)
  so, in total, the surface area \(S\) is: \[S = 2\pi r^2 + 2\pi r h\]

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• Step 2 (write the constraint in terms of the variables): The constraint is the fact that the can has to be able to hold \(330\) ml of soda. This tells us that the cylinder's volume, which we can call \(V\) must be \(330\) ml.

  We're told, in the question, that \(330\) ml corresponds to \(330\) \(cm^3\). So the contraint is: \[V = 330\]

  Given that a cylinder, of radius \(r\) and height \(h\) has volume: \[V = h.\pi r^2\] We can now write the constraint for our soda can in terms of the variables (\(r\) and \(h\)): \[h.\pi r^2 = 330\]

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• Step 3 (use the constraint to eliminate one of the variables from the quantity we're optimizing): rearranging our expression for the constraint: \[h.\pi r^2 = 330\] we can make \(h\) the subject: \[h = \frac{330}{\pi r^2}\] Now we can replace \(h\) in our expression for the soda can's surface area, \(S = 2\pi r^2 + 2\pi r h\), to express \(S\) in terms of \(r\) only: \[\begin{aligned} S & = 2\pi r^2 + 2\pi r \times \frac{330}{\pi r^2} \\ & = 2 \pi r^2 + \frac{{660}\pi r}{\pi r^2} \\ S & = 2\pi r^2 + \frac{660}{r} \end{aligned}\] which we can write as a function of \(r\): \[S(r) = 2\pi r^2 + \frac{660}{r}\]

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• Step 4 (use calculus to optimize): to find the can's minimum surface area, we turn to calculus and start by differentiating \(S(r)\) with respect to \(r\): \[S'(r) = 4\pi r - \frac{660}{r^2}\] We now solve \(S'(r)= 0\), that's: \[\begin{aligned} & 4\pi r - \frac{660}{r^2} = 0 \\ & 4 \pi r = \frac{660}{r^2} \\ & 4 \pi r^3 = 660 \\ & r^3 = \frac{660}{4\pi} = \frac{165}{\pi} \\ & r = \sqrt[3]{\frac{165}{\pi}} \approx 3.74 \ \text{cm} \end{aligned}\]

Important Note: although it's quite clear that this value of \(r\) can only correspond to a minimum surface area (since, in theory, without any constraints on the amount of material we can use: there is no limit to how big we can make the can and therefore no maximum surface area) if needs be: we can confirm that this value of \(r\) does indeed correspond to a minimum surface area using the second derivative test.

To find the second derivative, \(S''(r)\) we differentiate \(S'(r) = 4\pi r - \frac{660}{r^2}\) with respect to \(r\): \[S''(r) = 4 \pi + \frac{1320}{r^3}\] So when \(r = \sqrt[3]{\frac{165}{\pi}}\) that's: \[S''\begin{pmatrix}\sqrt[3]{\frac{165}{ \pi}} \end{pmatrix} = 4\pi + \frac{1320}{\begin{pmatrix} \sqrt[3]{\frac{165}{ \pi}} \end{pmatrix}^3}\] That's: \[S''\begin{pmatrix}\sqrt[3]{\frac{165}{\pi}} \end{pmatrix} = 4\pi + \frac{1320}{\frac{165}{\pi}}\] Remembering that all that matters, when using the second derivative test is the sign or the second derivative, it's quite clear that \(S''\begin{pmatrix}\sqrt[3]{\frac{165}{\pi}} \end{pmatrix}\) is positive (since both terms in the sum are positive). We can therefore state: \[S''\begin{pmatrix}\sqrt[3]{\frac{165}{\pi}} \end{pmatrix} > 0\] and that the second derivative test allows us to state that when \(r = \sqrt[3]{\frac{165}{\pi}} \approx 3.74 \ \text{cm}\) the surface area, \(S(r)\), reaches a minimum value.

The corresponding height, \(h\), of the can can/should then be found using the expression we wrote at the beginning of Step 3, \(h = \frac{330}{\pi r^2}\), using \(r = \sqrt[3]{\frac{165}{\pi}}\), that's: \[h = \frac{330}{\pi \begin{pmatrix} \sqrt[3]{\frac{165}{ \pi}}\end{pmatrix}^2} \approx 7.49 \ \text{cm}\]

Finally we can state that the optimum dimensions to minimize the surface area of the soda can are:

• radius : \(r = 3.74\) cm
• height : \(h= 7.49\) cm

The shortest distance from the point \(\begin{pmatrix}2,0 \end{pmatrix}\) to the curve \(y=\sqrt{x}\) is: \[D = \frac{\sqrt{7}}{2} \approx 1.32\] That's \(1.32\) units of length.

• Step 1 (find an expression for the quantity we need to optimize, in term of the problem's variables): The quantity we're tyring to opyimize (minimize in this case) is the distance from the point \(\begin{pmatrix}2,0\end{pmatrix}\) to the curve \(y=\sqrt{x}\), which we can call \(D\); another way of thinking of this problem is that we're looking for the point on the curve \(y=\sqrt{x}\) that is closest to the point \(\begin{pmatrix}2,0\end{pmatrix}\).
  
  The distance from the point \(\begin{pmatrix}2,0\end{pmatrix}\) to any point \(\begin{pmatrix}x,y\end{pmatrix}\) in the \(xy\)-plane (whether that point be on the curve \(y=\sqrt{x}\) or not) is: \[D = \sqrt{\begin{pmatrix}x-2\end{pmatrix}^2 + \begin{pmatrix}y-0\end{pmatrix}^2}\] That's: \[D = \sqrt{\begin{pmatrix}x-2\end{pmatrix}^2 + y^2}\]

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  Note: The formula we used was the formula for the distance beetween 2 points. It states that the distance, \(D\), between 2 points \(P\begin{pmatrix}x_1,y_1\end{pmatrix}\) and \(Q\begin{pmatrix}x_2,y_2\end{pmatrix}\) is: \[D = \sqrt{\begin{pmatrix}x_2-x_1\end{pmatrix}^2 + \begin{pmatrix}y_2-y_1\end{pmatrix}^2}\]

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• Step 2 (write the constraint in terms of the variables): in this case the only thing stopping us from making the distance from the point \(\begin{pmatrix}2,0\end{pmatrix}\) equal to zero is the fact that the other point has to lie along the curve \(y=\sqrt{x}\); that is the constraint (the fact that \(y\) must equal to \(\sqrt{x}\)). So, the constraint is "simply": \[y=\sqrt{x}\]

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• Step 3 (use the constraint to eliminate one of the variables from the quantity we're optimizing): Our expressing for the constraint is: \[y=\sqrt{x}\] Replacing the \(y\) inside our expression for the distance, \(D=\sqrt{\begin{pmatrix}x-2\end{pmatrix}^2 + y^2}\), we obtain: \[D = \sqrt{\begin{pmatrix}x-2\end{pmatrix}^2 + \begin{pmatrix}\sqrt{x}\end{pmatrix}^2}\] That's: \[D = \sqrt{\begin{pmatrix}x-2\end{pmatrix}^2 + x}\] Expanding \(\begin{pmatrix}x-2\end{pmatrix}^2\): \[D = \sqrt{x^2 - 4x + 4 + x}\] gathering like terms we find: \[D(x) = \sqrt{x^2 - 3x + 4}\] We can now see that the distance is only a function of \(x\), \(D(x)\), and that we have successfully eliminated the variable \(y\).

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• Step 4 (use calculus to optimize): we're looking for the minimum value of \(D(x)\). We start by differentiating \(D(x)\) with respect to \(x\): \[D'(x) = \frac{2x - 3}{2\sqrt{x^2-3x+4}}\] We now solve \(D'(x)=0\), that's: \[\frac{2x - 3}{2\sqrt{x^2-3x+4}} = 0\] Like any other fraction, \(\frac{2x - 3}{2\sqrt{x^2-3x+4}}\) can only equal to \(0\) if the numerator equals zero. So we solve: \[2x-3 = 0\] which leads to: \[x = \frac{3}{2}\] At this value of \(x\), the function \(D(x)\) reaches a stationary point. The \(y\) coordinate is: \[y = \sqrt{\frac{3}{2}}\] And so the point \(\begin{pmatrix} \frac{3}{2}, \sqrt{\frac{3}{2}} \end{pmatrix}\) is a stationary point. But is it a minimum, a maximum, or even a horizontal point of inflection ?

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  Note: given the context, it is quite clear that this stationary point can't be a maximum. Indeed, since the curve \(y=\sqrt{x}\) goes-on and increases infinitely, it isn't possible to have a maximum distance between \(\begin{pmatrix}2,0 \end{pmatrix}\) and a point on the curve.
  Furthermore, since we were asked to find the "minimum distance" and we only found one stationary point it seems reasonable to assume that this stationary point mut be a minimum.

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  But if we need to provide rigorous mathematical proof then we can study the sign of the derivative \(D'(x)\) on either side of the stationary point.
  Remember we had found: \[D'(x) = \frac{2x - 3}{2\sqrt{x^2-3x+4}}\] Notice that the denominator \(2\sqrt{x^2 - 3x - 4}\) will always be positive. The numerator \(2x-3\) is therefore what dictates the sign of the derivative. We can therefore construct a sign table for \(D'(x)\) "simply" by studying the esign of \(2x-3\): We can see from the sign table that the stationary point at \(\begin{pmatrix} \frac{3}{2}, \sqrt{\frac{3}{2}} \end{pmatrix}\) is a minimum.
  Since \(D(x)\) corresponds to the distance from the point \(\begin{pmatrix}2,0\end{pmatrix}\) to a point \(\begin{pmatrix}x,y\end{pmatrix}\) on thee curve \(y=\sqrt{x}\): we can conclude that the point \(\begin{pmatrix} \frac{3}{2}, \sqrt{\frac{3}{2}} \end{pmatrix}\) is the closest to \(\begin{pmatrix}2,0\end{pmatrix}\).
  To find what the shortest distance is, we can plug-in the value \(x=\frac{3}{2}\) into the expression we found for \(D(x)\) at the end of Step 3, \(D(x) = \sqrt{x^2 - 3x + 4}\). That's: \[\begin{aligned} D\begin{pmatrix} \frac{3}{2} \end{pmatrix} & = \sqrt{\begin{pmatrix} \frac{3}{2} \end{pmatrix}^2 - 3\begin{pmatrix} \frac{3}{2} \end{pmatrix} + 4} \\ & = \sqrt{ \frac{9}{4}  - \frac{9}{2}  + 4} \\ & = \sqrt{\frac{7}{4}} \\ D\begin{pmatrix} \frac{3}{2} \end{pmatrix} & = \frac{\sqrt{7}}{2} \quad \begin{pmatrix} \approx 1.32 \end{pmatrix} \end{aligned}\] Finally, we can state that the shortest distance from the point \(\begin{pmatrix} 2, 0 \end{pmatrix}\) to the curve \(y=\sqrt{x}\) is \(1.32\) units of length (rounded to 2 deecimal places).