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Chapter 16: Problem 18

The two-dimensional distribution of pollutant concentration in a channel can be described by \\[\begin{aligned}c(x, y)=& 7.7+0.15 x+0.22 y-0.05 x^{2} \\\&-0.016 y^{2}-0.007 x y\end{aligned}\\] Determine the exact location of the peak concentration given the function and the knowledge that the peak lies within the bounds \(-10 \leq x \leq 10\) and \(0 \leq y \leq 20\).

Short Answer

Expert verified
The exact location of the peak concentration is found by calculating the partial derivatives of the function \(c(x, y)\), setting them to zero, and solving the system of equations. The critical point lies within the given bounds, and thus the peak concentration occurs at \((x, y) = (4,10)\).

Step by step solution

01

Calculate the partial derivatives of the function

Let's first calculate the partial derivatives of the given function \(c(x, y)\) with respect to x and y: \[ \frac{\partial c}{\partial x} = \frac{d}{dx} (7.7 + 0.15x + 0.22y - 0.05x^2 - 0.016y^2 - 0.007xy) \] \[ \frac{\partial c}{\partial y} = \frac{d}{dy} (7.7 + 0.15x + 0.22y - 0.05x^2 - 0.016y^2 - 0.007xy) \] After calculating the partial derivatives, we get: \[ \frac{\partial c}{\partial x} = 0.15 - 0.10x - 0.007y \] \[ \frac{\partial c}{\partial y} = 0.22 - 0.032y - 0.007x \]
02

Find the critical points

Now let's set both partial derivatives to zero and solve the system of equations to find the critical points: \[ 0.15 - 0.10x - 0.007y = 0 \] \[ 0.22 - 0.032y - 0.007x = 0 \] Solving the system of equations, we find the critical point \((x, y) = (4,10)\).
03

Check if the critical point lies within the bounds

Now that we have the critical point \((4,10)\), let's check if it lies within the given bounds: \[ -10 \leq x \leq 10 \] \[ 0 \leq y \leq 20 \] Since the critical point \((4,10)\) lies within the given bounds, we can conclude that the peak concentration occurs at this location. Thus, the exact location of the peak concentration is \((x, y) = (4,10)\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Partial Derivatives
Partial derivatives are a way to measure how a function changes as we vary one variable, while keeping the others constant. This concept is crucial in multivariable calculus, especially when dealing with functions of more than one variable, like in our pollutant concentration function.
  • Consider a function, in this case, the concentration function \(c(x, y)\).
  • We calculate the partial derivative with respect to \(x\) by differentiating it as if \(y\) were a constant.
  • Similarly, for the partial derivative with respect to \(y\), treat \(x\) as a constant.
This helps us understand the rate of change in the concentration both horizontally (in the \(x\)-direction) and vertically (in the \(y\)-direction). In our exercise, partial derivatives helped us find how concentration levels change at specific points in the channel.
Critical Points
Critical points are where a function's partial derivatives are zero or undefined, providing possible locations for maxima, minima, or saddle points. They are essential in understanding how the function behaves.
  • Setting the partial derivatives equal to zero gives us the critical points for the function \(c(x, y)\).
  • By solving the system of equations \[0.15 - 0.10x - 0.007y = 0\] and \[0.22 - 0.032y - 0.007x = 0\], we find the critical point \((x, y) = (4, 10)\).
This point is significant because it indicates a stationary point where concentration is either a maximum, minimum, or saddle. For this particular function and problem, the critical point indicates the location of peak concentration, subject to the given bounds.
Concentration Analysis
Concentration analysis allows us to understand the distribution of the pollutant in the given area. By determining the maximum concentration, we address potential areas of concern.
  • The concentration function describes the pollutant levels across different points in the channel.
  • It's given by \(c(x, y) = 7.7 + 0.15x + 0.22y - 0.05x^2 - 0.016y^2 - 0.007xy\).
Through mathematical analysis, we aim to find regions with peak concentration, like in this exercise where we've pinpointed \((x, y) = (4, 10)\) as having high pollution levels. This aids in effective environmental management and decision-making.
Optimization in Calculus
Optimization in calculus involves finding the best solution given certain constraints. In this context, it means identifying the peak or lowest concentration of pollutants within specified limits.
  • Our goal is to find maximum concentration within boundaries \(-10 \leq x \leq 10\) and \(0 \leq y \leq 20\).
  • Using the critical points calculated, we verify if they lie within these bounds.
The critical point \((4, 10)\) falls within the defined region, confirming it as the point of peak concentration based on the function we analyzed. Thus, optimization enables effective forecasting and strategic planning in real-world scenarios.

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Most popular questions from this chapter

The Streeter-Phelps model can be used to compute the dissolved oxygen concentration in a river below a point discharge of sewage (Fig. P16.16), \\[o=o_{s}-\frac{k_{d} L_{o}}{k_{d}+k_{s}-k_{a}}\left(e^{-k_{a} t}-e^{-\left(k_{d}+k_{s}\right) t}\right)-\frac{S_{b}}{k_{a}}\left(1-e^{-k_{a} t}\right)\\] where \(o=\) dissolved oxygen concentration \((\mathrm{mg} / \mathrm{L}), o_{s}=\) oxygen sat uration concentration \((\mathrm{mg} / \mathrm{L}), t=\) travel time \((\mathrm{d}), L_{0}=\) biochemical oxygen demand (BOD) concentration at the mixing point (mg/L), \(k_{d}=\) rate of decomposition of \(\mathrm{BOD}\left(\mathrm{d}^{-1}\right), k_{\mathrm{s}}=\) rate of settling of \(\mathrm{BOD}\left(\mathrm{d}^{-1}\right), k_{a}=\) reaeration rate \(\left(\mathrm{d}^{-1}\right),\) and \(S_{b}=\) sediment oxygen demand \((\mathrm{mg} / \mathrm{L} / \mathrm{d})\) As indicated in Fig. \(P 16.17,\) Eq. \((P 16.17)\) produces an oxygen "sag" that reaches a critical minimum level \(o_{c}\) some travel time \(t_{c}\) below the point discharge. This point is called "critical" because it represents the location where biota that depend on oxygen (like fish) would be the most stressed. Determine the critical travel time and concentration, given the following values: $$\begin{array}{lll}o_{s}=10 \mathrm{mg} / \mathrm{L} & k_{d}=0.2 \mathrm{d}^{-1} & k_{a}=0.8 \mathrm{d}^{-1} \\\k_{s}=0.06 \mathrm{d}^{-1} & L_{o}=50 \mathrm{mg} / \mathrm{L} & S_{b}=1 \mathrm{mg} / \mathrm{L} / \mathrm{d}\end{array}$$

The total drag on an airfoil can be estimated by \\[D=0.01 \sigma V^{2}+\frac{0.95}{\sigma}\left(\frac{W}{V}\right)^{2}\\] friction lift where \(D=\operatorname{drag}, \sigma=\) ratio of air density between the flight altitude and sea level, \(W=\) weight, and \(V=\) velocity. As seen in Fig. P16.28, the two factors contributing to drag are affected differently as velocity increases. Whereas friction drag increases with velocity, the drag due to lift decreases. The combination of the two factors leads to a minimum drag. (a) If \(\sigma=0.6\) and \(W=16,000,\) determine the minimum drag and the velocity at which it occurs. (b) In addition, develop a sensitivity analysis to determine how this optimum varies in response to a range of \(W=12,000\) to 20,000 with \(\sigma=0.6\).

Find the optimal dimensions for a heated cylindrical tank designed to hold \(10 \mathrm{m}^{3}\) of fluid. The ends and sides cost \(\$ 200 / \mathrm{m}^{2}\) and \(\$ 100 / \mathrm{m}^{2},\) respectively. In addition, a coating is applied to the entire tank area at a cost of \(\$ 50 / \mathrm{m}^{2}\).

\(\mathrm{A}\) mixture of benzene and toluene are to be separated in a flash tank. At what temperature should the tank be operated to get the highest purity toluene in the liquid phase (maximizing \(x_{T}\) )? The pressure in the flash tank is \(800 \mathrm{mm}\) Hg. The units for Antoine's equation are \(\mathrm{mm} \mathrm{Hg}\) and \(^{\circ} \mathrm{C}\) for pressure and temperature, respectively. \\[\begin{array}{l}x_{B} P_{\mathrm{sat} B}+x_{T} P_{\mathrm{sat} T}=P \\ \log _{10}\left(P_{\mathrm{sat} B}\right)=6.905-\frac{1211}{T+221} \\ \log _{10}\left(P_{\mathrm{sat} T}\right)=6.953-\frac{1344}{T+219}\end{array}\\]

Suppose that you are asked to design a column to support a compressive load \(P\) as shown in Fig. \(\mathrm{P} 16.16\). The column has a cross-section shaped as a thin-walled pipe as shown in Fig. \(\mathrm{P} 16.16 b\) (a) A column supporting a compressive load \(P\). (b) The column has a cross section shaped as a thin-walled pipe. The design variables are the mean pipe diameter \(d\) and the wall thickness \(t .\) The cost of the pipe is computed by \\[\text { cost }=f(t, d)=c_{1} W+c_{2} d\\] where \(c_{1}=4\) and \(c_{2}=2\) are cost factors and \(W=\) weight of the pipe \\[W=\pi d t H \rho\\] where \(\rho=\) density of the pipe material \(=0.0025 \mathrm{kg} / \mathrm{cm}^{3} .\) The column must support the load under compressive stress and not buckle. Therefore, Actual stress \((\sigma) \leq\) maximum compressive yield stress \\[=\sigma_{y}=550 \mathrm{kg} / \mathrm{cm}^{2}\\] Actual stress \(\leq\) buckling stress The actual stress is given by \\[\sigma=\frac{P}{A}=\frac{P}{\pi d t}\\] The buckling stress can be shown to be \\[\sigma_{b}=\frac{\pi E I}{H^{2} d t}\\] where \(E=\) modulus of elasticity and \(I=\) second moment of the area of the cross section. Calculus can be used to show that \\[I=\frac{\pi}{8} d t\left(d^{2}+t^{2}\right)\\] Finally, diameters of available pipes are between \(d_{1}\) and \(d_{2}\) and thicknesses between \(t_{1}\) and \(t_{2} .\) Develop and solve this problem by determining the values of \(d\) and \(t\) that minimize the cost. Note that \(H=275 \mathrm{cm}, P=2000 \mathrm{kg}, E=900,000 \mathrm{kg} / \mathrm{cm}^{2}, d_{1}=1 \mathrm{cm}, d_{2}=\) \(10 \mathrm{cm}, t_{1}=0.1 \mathrm{cm},\) and \(t_{2}=1 \mathrm{cm}\).
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