In each of Problems 7 through 11, determine whether the members of the given set of vectors are linearly independent. If they ar
e linearly dependent, find a linear relation among then. The vectors are written as row vectors to save space but may be considered as column vectors that is, the transposes of the given vectors may be used instead of the vectors themselves. x^(1) = (1, 1, 0), x^(2) = (0, 1, 1), x^(3) = (1, 0, 1) x^(1) = (2, 1, 0) x^(2) = (0, 1, 0), x^(3) = (-1, 2, 0) x^(1) = (1, 2, 2, 3), x^(2) = (-1, 0, 3, 1), x^(3) = (-1, 2, 0) x^(4) = (-3, t-13) x^(1) = (1, 2, -1, 0), x^(2) = (2, 3, 1, -1), x^(3) = (-1, 0, 2, 2), x^(4) = (3, -1, 1, 3) Suppose that each of the vectors x^(1), ..., x^(m) has n components, where n < m. x^(1), ...x^(m) are linearly dependent. In each of Problems 13 and 14, determine whether the member of the given of vectors linearly independent for -infinity < t < infinity. If they are linearly dependent, find the linear among them. As in Problems 7 through 11, the vectors are written as row vectors to save space x^(1) (t) = (e^-t, 2e^-t), x^(2) (t) = (e6-t, e^(-t), x^(3) (t) = (3e^-t, 0) x^(1) (t) = (2 sin t, sin t), x^(2) (t) = (sin t, 2 sin t)
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Answer:
26 ft square by 13 ft high
Step-by-step explanation:
The tank will have minimum surface area when opposite sides have the same total area as the square bottom. That is, their height is half their width. This makes the tank half a cube. Said cube would have a volume of ...
2·(8788 ft^3) = (26 ft)^3
The square bottom of the tank is 26 ft square, and its height is 13 ft.
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<em>Solution using derivatives</em>
If x is the side length of the square bottom, the height is 8788/x^2 and the area is ...
x^2 + 4x(8788/x^2) = x^2 +35152/x
The derivative of this is zero when area is minimized:
2x -35152/x^2 = 0
x^3 = 17576 = 26^3 . . . . . multiply by x^2/2, add 17576
x = 26
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As the attached graph shows, a graphing calculator can also provide the solution.
