Find two positive numbers that satisfy the given requirements.

The product is 147 and the sum of the first number plus 3 times the second number is a minimum.

a*b = 147

147 = 3*7*7

So, the only choices for a*b are

3*49
49*3
7*21
21*7
1*147
147*1

You can figure which gives minimum a+3b

To find two positive numbers that satisfy the given requirements, we need to solve the problem mathematically. Let's denote the first number as "x" and the second number as "y".

According to the requirements, the product of the two numbers is 147: x * y = 147.

We also need to minimize the sum of the first number plus 3 times the second number: S = x + 3y.

To solve this problem, we can use the method of substitution. We can rewrite the equation for the product and solve for one variable in terms of the other.

x * y = 147
y = 147/x

Now, substitute the value of y in terms of x into the equation for the sum:

S = x + 3y
S = x + 3(147/x)
S = x + 441/x

The next step is to find the minimum value of the sum. To do that, we can take the derivative of S with respect to x and set it equal to zero.

dS/dx = 1 - 441/x^2
0 = 1 - 441/x^2
441/x^2 = 1
x^2 = 441
x = √441
x = 21

Now that we have the value of x, we can substitute it back into the equation for the product to find y:

y = 147/x
y = 147/21
y = 7

Therefore, the two positive numbers that satisfy the given requirements are x = 21 and y = 7.