Question

Find two positive numbers satisfying the given requirements. The product is 432 and the sum of the first plus three times the second is a minimum.

Answer 1

36 and 12

Explanation:

Let the two positive integers be x and y.

If their product is 432, then

xy = 432 ......... 1

Also if the sum of the first plus three times the second is a minimum, then;

p(x) = x + 3y

From 1;

y = 432/x ..... 3

Substitute 3 into 2;

p(x) = x+3y

p(x)= x + 3(432/x)

p(x) = x + 1296/x

Since the expression is at minimum when dp(x)/dx = 0

dp/dx = 1 + (-1296)/x²

dp/dx = 1 -1296/x²

0 = 1 -1296/x²

0 = (x²-1296)/x²

cross multiply

0 = x²-1296

x² = 1296

x = √1296

x = 36

Since xy = 432

36y = 432

y = 432/36

y = 12

Hence the two positive numbers are 36 and 12