Question

Let a and b be two positive numbers. If 2a + 3b=6 then the maximum product of these a and b is:

Answer 1

Answer:

a= 3/2

b= 1

a*b= 3/2

Step-by-step explanation:

For getting this problem we need to think in terms of only one unknown. Lets start by looking at our equation:

2a + 3b=6

No, we need to see what happens to a*b. Lets get the value of one of these unknowns in terms of the other, this is, a in terms of b or b in terms of a. Here, I will get a in terms of b (if you like, do the other way and compare):

2a + 3b=6

Subtract 3b in both sides:

2a = 6 - 3b

Now divide both sides by 2 to get a alone:

a = (6 - 3b)/2

a = 3 - 3/2 b

Now lets try to multiply a by b, using this a we found above:

a*b = (3 - 3/2 b)*b

Using distributive:

a*b = 3b - 3/2 b^2

So, we have an expression for a*b that depends only on b. Notice that this expression is a parabola with negative coefficient on the square term, what makes it a negative parabola that MUST have a maximum value. So, we have an expression for a*b that can be maximized, so we can find the maximum of a*b by the derivative of the expression. Lets derive in b:

(3b - 3/2 b^2)' = (3b)' - (3/2 b^2)' = 3 - 3b

So, the derivative equal to 0 gives us the maximum:

3 - 3b = 0

Suming 3b in both sides:

3 = 3b

Dividing by 3:

1 = b

So, we maximize our expression a*b when b is equal to 1. Now, we can replace it on a = 3 - 3/2 b to find b:

a = 3 - 3/2(1) = 3 - 3/2 = 3/2

Thus, a is equal to 3/2 and the product a*b is maximized when:

a = 3/2

b = 1

And the product is a*b= (3/2) * 1 = 3/2