Mike D. answered • 11/12/21

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Suppose the three fenced sides are w, w, l (with l parallel to the river)

Area = wl = 5000

L = length of fence = 2w + l = 2w + (5000/w)

dL/dw = 2 - (5000/w^{2})

d^{2}L / dw^{2} = 10000 / w^{3}

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To find max/min L, dL/dw = 0 giving 2w^{2} = 5000, w^{2} = 2500, w = 50

When w = 50, d^{2}L/dw^{2} > 0, showing w=50 is a minimum

wl = 5000 so l = 5000 / 50 = 100

dimensions are width 50, length 100, total fence length 200.