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The maximum value of a function on a line in space Does the function \(f(x, y, z)=x y z\) have a maximum value on the line\(x=20-t, y=t, z=20 ?\) If so, what is it? Give reasons for your answer. (Hint: Along the line, \(w=f(x, y, z)\) is a differentiable function of \(t . )\)

Short Answer

Expert verified
The maximum value is 2000 at \(t = 10\).

Step by step solution

01

Express Function in Terms of Parameter

Substitute the parametric equations of the line into the function \(f(x, y, z) = x \, y \, z\). This gives us \(f(x(t), y(t), z) = (20-t)\, t \, 20\). The simplified function in terms of \(t\) is \(w(t) = 20(20t - t^2) = 400t - 20t^2\).
02

Find the derivative of the function

Calculate the derivative of \(w(t) = 400t - 20t^2\) with respect to \(t\). The first derivative \(w'(t)\) is \(w'(t) = 400 - 40t\).
03

Determine critical points

Set the derivative \(w'(t) = 400 - 40t\) equal to zero to find the critical points: \(400 - 40t = 0\). Solve for \(t\), resulting in \(t = 10\).
04

Evaluate the function at critical points

Substitute \(t = 10\) into \(w(t) = 400t - 20t^2\) to find the maximum value: \(w(10) = 400(10) - 20(10)^2 = 4000 - 2000 = 2000\).
05

Analyze the function behavior for verification

Check the second derivative, \(w''(t) = -40\), which is negative, indicating \(w(t)\) is concave down, confirming a local maximum at \(t = 10\).

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Key Concepts

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

Maxima and Minima
In the realm of calculus, finding the maxima and minima of a function is essential for understanding its behavior. These points mark where the function reaches its highest or lowest value, respectively. They often represent important features in various disciplines, such as optimization problems in economics or engineering. To determine these points, one typically investigates where the derivative of the function equals zero. This is because a slope of zero signifies a turning point, such as a peak or a trough.

Once you find the critical points, it's crucial to check their nature. A helpful way to confirm whether these are maxima or minima is by taking the second derivative. If the second derivative at a point is negative, it's a local maximum; if positive, it's a local minimum. In our function, the second derivative test revealed a negative value, confirming the presence of a local maximum at the found critical point. These techniques ensure a thorough understanding of function behavior across its domain.
Differential Calculus
Differential calculus is all about the rate of change. By taking derivatives, we get an insight into how a function changes as its variables change. When assessing a function to find its maxima or minima, derivatives guide us in finding those critical points where the function changes direction.

In our exercise, after expressing the given function in terms of a single variable, differential calculus helped find when the slope of the curve became zero. This was achieved by deriving the function with respect to the variable and setting the result equal to zero. The solution focused on this derivative, leading to the discovery of the critical point.

Additionally, by evaluating the second derivative, we gain further insights into the function's concavity, telling us whether we're looking at a peak or a valley. This blend of first and second derivatives offers a powerful toolkit for understanding and optimizing functions, which is frequently applied in countless real-world scenarios.
Functions of Several Variables
Functions of several variables are built to handle more complex scenarios than functions with just one variable. In such functions, the idea is to explore how multiple variables interplay to affect outcomes. For example, given a function of the form \( f(x, y, z) = xyz \), each of these variables contributes to the final value of the function.

In this exercise, the function is initially in three variables. By using a parametric representation, we brought it down to a single variable problem where optimization becomes more straightforward. The relationship among multiple variables was simplified, allowing us to tackle it using the intuitive methods of single-variable calculus.

This approach highlights the convenience and necessity of understanding functions in multiple variables, especially when considering constraints or specific paths like lines in space. Such techniques are essential in multivariable calculus, playing a significant role in fields like economics, physics, and computer graphics, where multiple factors must be simultaneously considered.

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Most popular questions from this chapter

To find the extreme values of a function \(f(x, y)\) on a curve \(x=x(t), y=y(t),\) we treat \(f\) as a function of the single variable \(t\) and use the Chain Rule to find where \(d f / d t\) is zero. As in any other single-variable case, the extreme values of \(f\) are then found among the values at the a. critical points (points where \(d f / d t\) is zero or fails to exist), and b. endpoints of the parameter domain. Find the absolute maximum and minimum values of the following functions on the given curves. Functions: a. \(f(x, y)=2 x+3 y \quad\) b. \(g(x, y)=x y\) c. \(h(x, y)=x^{2}+3 y^{2}\) Curves: i. The semi-ellipse \(\left(x^{2} / 9\right)+\left(y^{2} / 4\right)=1, \quad y \geq 0\) ii. The quarter ellipse \(\left(x^{2} / 9\right)+\left(y^{2} / 4\right)=1, \quad x \geq 0, \quad y \geq 0\) Use the parametric equations \(x=3 \cos t, y=2 \sin t\)

Use a CAS to perform the following steps implementing the method of Lagrange multipliers for finding constrained extrema: a. Form the function \(h=f-\lambda_{1} g_{1}-\lambda_{2} g_{2},\) where \(f\) is the function to optimize subject to the constraints \(g_{1}=0\) and \(g_{2}=0 .\) b. Determine all the first partial derivatives of \(h\) , including the partials with respect to \(\lambda_{1}\) and \(\lambda_{2},\) and set them equal to \(0 .\) c. Solve the system of equations found in part (b) for all the unknowns, including \(\lambda_{1}\) and \(\lambda_{2} .\) d. Evaluate \(f\) at each of the solution points found in part (c) and select the extreme value subject to the constraints asked for in the exercise. Minimize \(f(x, y, z)=x^{2}+y^{2}+z^{2}\) subject to the constraints \(x^{2}-x y+y^{2}-z^{2}-1=0\) and \(x^{2}+y^{2}-1=0\)

In Exercises \(51-56,\) find the limit of \(f\) as \((x, y) \rightarrow(0,0)\) or show that the limit does not exist. $$ f(x, y)=\cos \left(\frac{x^{3}-y^{3}}{x^{2}+y^{2}}\right) $$

In Exercises \(25-30,\) find the linearization \(L(x, y)\) of the function at each point. $$ f(x, y)=e^{2 y-x} \text { at } \quad \text { a. }(0,0), \quad \text { b. }(1,2) $$

In Exercises \(51-56,\) find the limit of \(f\) as \((x, y) \rightarrow(0,0)\) or show that the limit does not exist. $$ f(x, y)=\tan ^{-1}\left(\frac{|x|+|y|}{x^{2}+y^{2}}\right) $$

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