In the answer space below, provide the larger of the two positive integers that add up to 60 and have the largest possible product.
30
think of a rectangle. The rectangle with the largest area is a square.
So, 30^2 = 900
Any other two numbers with sum=60 will have a smaller product. They will be (30-k) and (30+k), whose product is 30^2 - k^2
To find the positive integers that add up to 60 and have the largest possible product, we can use the concept of maximizing the product of two numbers given their sum.
Let's assume the two positive integers are x and y. Since they add up to 60, we have the equation:
x + y = 60
To find the largest possible product, we need to maximize the value of xy.
To proceed, we can try different values of x and calculate the corresponding value of y. Let's start with x = 30 (which is half of 60):
x = 30
y = 60 - x
y = 60 - 30
y = 30
So, if x = 30, then y = 30.
The product of x and y is obtained by multiplying them:
xy = 30 * 30
xy = 900
Let's also try another value, x = 31:
x = 31
y = 60 - x
y = 60 - 31
y = 29
So, if x = 31, then y = 29.
The product of x and y is obtained by multiplying them:
xy = 31 * 29
xy = 899
Comparing the two products, we see that the largest product is 900, obtained when x = y = 30.
Therefore, the larger of the two positive integers that add up to 60 and have the largest possible product is 30.
To find the larger of the two positive integers that add up to 60 and have the largest possible product, we need to consider the concept of maximizing the product of two numbers given their sum.
Let's assume the two numbers are x and y. We know that x + y = 60. We need to find the values of x and y that will maximize the product xy.
To solve this, we can rewrite the equation x + y = 60 as y = 60 - x. Now, we can express the product of x and y in terms of x alone: P = x(60 - x).
To find the maximum value of P, we can take the derivative of P with respect to x and set it equal to zero. This will give us the critical points where P is maximized.
Taking the derivative of P = x(60 - x) with respect to x, we get dP/dx = 60 - 2x.
Setting dP/dx = 0, we have 60 - 2x = 0. Solving this equation, we find x = 30.
Since we are looking for positive integers, and x = 30, we can substitute this value back into the equation y = 60 - x to find y: y = 60 - 30 = 30.
Therefore, the two positive integers that add up to 60 and have the largest possible product are 30 and 30. So, the larger of the two numbers is 30.