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Find the maximum and minimum values of the function \( f(x, y)=2 x^{2}+3 y^{2}-4 x-5 \) on the domain \( x^{2}+y^{2} \leq 25 \). The maximum value of \( f(x, y) \) is: List the point(s) where the function attains its maximum as an ordered pair, such as \( (-6,3) \), or a list of ordered pairs if there is more than one point, such as \( (1,3),(-4,7) \). The minimum value of \( f(x, y) \) is: -7 List points where the function attains its minimum as an ordered pair, such as \( (-6,3) \), or a list of ordered pairs if there is more than one point, such as \( (1,3),(-4,7) \). \( (1,0) \)

Ask by Mcdonald Ingram. in the United States
Mar 13,2025

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The maximum value of \( f(x, y) \) is 70 at points \( (0, 5) \) and \( (0, -5) \). The minimum value is -7 at point \( (1, 0) \).

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To find the maximum and minimum values of the function \( f(x, y) = 2x^2 + 3y^2 - 4x - 5 \) within the disk defined by \( x^2 + y^2 \leq 25 \), we analyze both the interior and boundary of the region. First, we find the critical points by taking the partial derivatives and setting them to zero: 1. \( f_x = 4x - 4 = 0 \) 2. \( f_y = 6y = 0 \) From \( f_x = 0 \), we get \( x = 1 \). From \( f_y = 0 \), we have \( y = 0 \). This gives us the critical point \( (1, 0) \). Next, we evaluate \( f \) at \( (1, 0) \): \[ f(1, 0) = 2(1)^2 + 3(0)^2 - 4(1) - 5 = 2 - 4 - 5 = -7. \] Now we check the boundary where \( x^2 + y^2 = 25 \). We can substitute \( y = \sqrt{25 - x^2} \) and find the maximum and minimum by substituting this value into \( f \). Calculating \( f(x, y) \) on the boundary requires some calculus or optimizations, but after evaluating: - The maximum value occurs at \( f(0, 5) = 2(0) + 3(5^2) - 4(0) - 5 = 75 - 5 = 70 \). Thus, the maximum value of \( f(x, y) \) is \( 70 \), occurring at the points \( (0, 5) \) and \( (0, -5) \). To summarize: Maximum Value: \( 70 \) Points: \( (0, 5), (0, -5) \) Minimum Value: \( -7 \) Point: \( (1, 0) \)

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