x y = 147 so x = 147/y
x + 3 y = a
147/y + 3 y = a
da/dy = 0 for max or min = 3 -147/y^2
y^2 = 147/3 = 49
y = 7
x = 147/7 = 21 so x = 21 and y = 7
Find two positive numbers that satisfy the given requirements: The product is 147 and the sum of the first number plus three times the second number is minimum
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