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Magazine Chapter 14: Problem 36
In Exercises \(31-38,\) find the absolute maxima and minima of the functions on the given domains. \(f(x, y)=48 x y-32 x^{3}-24 y^{2}\) on the rectangular plate \(0 \leq x \leq\) \(1,0 \leq y \leq 1\)
Short Answer
Expert verified
Max: \(f(0,0)=0\); Min: \(f(1,0)=-32\), \(f(0,1)=-24\).
Step by step solution
01
Understand the Problem
We need to find the absolute maxima and minima of the function \(f(x, y) = 48xy - 32x^3 - 24y^2\) on the domain \(0 \leq x \leq 1\) and \(0 \leq y \leq 1\). This means we need to find critical points and evaluate the function on the boundaries of the domain.
02
Find Critical Points
To find the critical points, we first find the partial derivatives \(f_x\) and \(f_y\). Calculate: \(f_x = \frac{\partial}{\partial x}(48xy - 32x^3 - 24y^2) = 48y - 96x^2\) and \(f_y = \frac{\partial}{\partial y}(48xy - 32x^3 - 24y^2) = 48x - 48y\). Set these derivatives equal to zero to find critical points: \(48y - 96x^2 = 0\) and \(48x - 48y = 0\). Solving these, we get \(y = 2x^2\) and \(x = y\). Substituting \(x = y\) into \(y = 2x^2\) results in \(x = 0\) or \(x = 1\). Yielding critical points \((0, 0)\) and \((1, 1)\).
03
Evaluate Function at Critical Points
Evaluate the function at critical points: \(f(0, 0) = 0\) and \(f(1, 1) = 48(1)(1) - 32(1)^3 - 24(1)^2 = -8\).
04
Evaluate Function on Boundaries
Evaluate the function on boundaries: For \(x=0\), \(f(0,y) = -24y^2\), which is minimized at \(y = 1\) to \(f(0, 1) = -24\). For \(x=1\), \(f(1,y) = 48y - 32 - 24y^2\), maximum analyzed via derivative, and values calculated at endpoints \(y=0\) and \(y=1\), giving \(f(1,0) = -32\) and \(f(1,1) = -8\). Similarly for \(y=0\), \(f(x,0) = -32x^3\), maximum at \(x=1\) gives \(f(1,0) = -32\). For \(y=1\), \(f(x,1) = 48x - 32x^3 - 24\), analyze end values; \(f(0,1)=-24\), \(f(1,1) = -8\).
05
Compare Values for Extremes
Compare all evaluated critical points and boundary values: \(f(0, 0) = 0\), \(f(1, 1) = -8\), \(f(0, 1) = -24\), \(f(1, 0) = -32\). The highest value is \(0\), and the lowest value is \(-32\). Thus, absolute maximum is at \((0,0)\) with value \(0\) and absolute minimum is at \((1,0)\) and \((0,1)\) with value \(-32\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Critical Points
Critical points are where the function's rate of change stops or changes direction. In multivariable calculus, these are the points on a surface where the gradient (a vector of partial derivatives) is zero or undefined.
Finding critical points is crucial because these points can indicate potential locations for local maxima, minima, or saddle points (areas that are neither maxima nor minima). To find these in the exercise, we calculate partial derivatives and set them to zero. This method helps identify where the function has no slope in any direction.
For a given function like our example, you solve the system of equations derived from setting the partial derivatives equal to zero. This exercise, we saw that the function reached critical points at \(0,0\) and \(1,1\). This process is essential, as solving for these points helps identify candidates for extrema.
Finding critical points is crucial because these points can indicate potential locations for local maxima, minima, or saddle points (areas that are neither maxima nor minima). To find these in the exercise, we calculate partial derivatives and set them to zero. This method helps identify where the function has no slope in any direction.
For a given function like our example, you solve the system of equations derived from setting the partial derivatives equal to zero. This exercise, we saw that the function reached critical points at \(0,0\) and \(1,1\). This process is essential, as solving for these points helps identify candidates for extrema.
Partial Derivatives
Partial derivatives are a way to understand how a function changes as each input variable changes, holding other variables constant. They're akin to regular derivatives but extend the concept to functions of multiple variables.
In our specific context, we calculated the partial derivatives of the function by taking the derivative of \(f(x, y) = 48xy - 32x^3 - 24y^2\) with respect to \(x\) and \(y\). This yields \(f_x = 48y - 96x^2\) and \(f_y = 48x - 48y\). These derivatives indicate the rate of change of the function concerning each variable.
Partial derivatives are foundational in determining critical points and provide insights into the function's behavior, especially when identifying directions of increase and decrease. This step highlights where the function doesn鈥檛 change and is essential to finding critical points that could be extrema.
In our specific context, we calculated the partial derivatives of the function by taking the derivative of \(f(x, y) = 48xy - 32x^3 - 24y^2\) with respect to \(x\) and \(y\). This yields \(f_x = 48y - 96x^2\) and \(f_y = 48x - 48y\). These derivatives indicate the rate of change of the function concerning each variable.
Partial derivatives are foundational in determining critical points and provide insights into the function's behavior, especially when identifying directions of increase and decrease. This step highlights where the function doesn鈥檛 change and is essential to finding critical points that could be extrema.
Extrema
Extrema refer to the maximum and minimum values of a function within its domain. These extremal points are critical in determining the function's highest peaks or lowest valleys.
In our exercise, finding the extrema involved evaluating the function at all critical points and along the boundaries of the specified domain \(0 \leq x \leq 1, \ 0 \leq y \leq 1\). It is critical to check the function鈥檚 behavior at the boundaries because extremum can also occur there, not just at the critical points.
After calculating the function values at critical points \(0,0\) and \(1,1\), and multiple evaluations along the boundary, we finally determine the absolute maximum and minimum. This exercise showed an absolute maximum at \(0,0\) with a value of 0, while the lowest occurred at \(1,0\) and \(0,1\) with values of \(-32\). This step consolidates the understanding of where the function is largest and smallest.
In our exercise, finding the extrema involved evaluating the function at all critical points and along the boundaries of the specified domain \(0 \leq x \leq 1, \ 0 \leq y \leq 1\). It is critical to check the function鈥檚 behavior at the boundaries because extremum can also occur there, not just at the critical points.
After calculating the function values at critical points \(0,0\) and \(1,1\), and multiple evaluations along the boundary, we finally determine the absolute maximum and minimum. This exercise showed an absolute maximum at \(0,0\) with a value of 0, while the lowest occurred at \(1,0\) and \(0,1\) with values of \(-32\). This step consolidates the understanding of where the function is largest and smallest.
