find two positive integers such that the sum of the first number and 4 times the second number is 1000 and the product of the numbers is as large as possible

Why did the two positive integers go on a diet? Because they wanted to multiply their potential and sum up to 1000 without any excess baggage!

So, let's find the perfect pair by maximizing their product. We can start by letting the first number be x and the second number be y. According to the information given, we have the following equation:

x + 4y = 1000

To maximize the product, we need to consider the fact that the two numbers should be as close together as possible. So, let's set y = x.

Substituting y = x into the equation, we get:

x + 4x = 1000
5x = 1000
x = 200

Now, we can find y by substituting x back into one of the equations:

y = x = 200

Therefore, the two positive integers that satisfy the conditions and maximize their product are x = 200 and y = 200.

I hope this mathematical journey brought a smile to your face!

To find two positive integers that satisfy the given conditions, we can start by assigning variables to the numbers. Let's call the first number "x" and the second number "y".

According to the problem, the sum of the first number and 4 times the second number is equal to 1000. We can express this as the equation:

x + 4y = 1000

Now, we want to find the values of x and y that will maximize the product of the numbers. The product of x and y can be represented by the equation:

xy

To maximize the product, we can use a basic rule of algebra which states that for any two positive integers whose sum is constant, their product is maximized when they are equal.

Since we can't set x = y in this case, we'll use the approximation that the numbers should be as close to each other as possible.

Let's solve the equation x + 4y = 1000 for x, keeping this approximation in mind:

x = 1000 - 4y

Now, substitute the value of x in the expression xy:

xy = (1000 - 4y)y = 1000y - 4y^2

To find the value of y that maximizes the product, we can take the derivative of xy with respect to y and set it equal to zero (to find the maximum point):

d(xy)/dy = 1000 - 8y = 0

Solving for y:

1000 - 8y = 0
8y = 1000
y = 125

Substitute the value of y back into the equation x + 4y = 1000 to find x:

x + 4(125) = 1000
x + 500 = 1000
x = 500

Therefore, the two positive integers are x = 500 and y = 125, which satisfy the given conditions and maximize their product.

To find two positive integers that satisfy the given conditions, we can solve the problem algebraically using a system of equations. Let's call the first number x and the second number y.

We are given two conditions:
1. The sum of the first number and 4 times the second number is 1000: x + 4y = 1000
2. The product of the numbers is as large as possible: maximize xy

To solve this system of equations, we can use substitution or elimination method. Let's use the substitution method:

1. Solve the first equation for x: x = 1000 - 4y
2. Substitute x into the second equation: (1000 - 4y) * y = maximize

Now we can find the equation for the product and maximize it:

P = (1000 - 4y) * y = 1000y - 4y^2

To maximize this equation, we need to find the vertex of the parabola formed by this equation. The x-coordinate of the vertex can be found using the formula: x = -b / (2a), where a = -4 and b = 1000. Substituting these values:

x = -1000 / (2 * (-4)) = -1000 / -8 = 125

So, to maximize the product, y = 125. We can substitute this value into any of the original equations to find x:

x = 1000 - 4y = 1000 - 4 * 125 = 1000 - 500 = 500

Therefore, the two positive integers that satisfy the conditions and maximize the product are x = 500 and y = 125.

let the second number be x,

let the first number be y
so y + 4x = 1000, ---> y = 1000-x

Product = xy
= x(1000-4x)
= 1000x - 4x^2

find d(product)/dx
set it equal to zero and solve for x
comes out very nice.