Mike D. answered 11/12/21
Effective, patient, empathic, math and science tutor
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/w2)
d2L / dw2 = 10000 / w3
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To find max/min L, dL/dw = 0 giving 2w2 = 5000, w2 = 2500, w = 50
When w = 50, d2L/dw2 > 0, showing w=50 is a minimum
wl = 5000 so l = 5000 / 50 = 100
dimensions are width 50, length 100, total fence length 200.