Find two numbers such that the sum of the first plus three times the second is 600 and their product is a maximum.
x + 3y=600
xy is a max. This is the problem statement.
let U be the function equal to xy.
U=xy = y(600-3y)
Now, maximize U (take derivative, set to zero), and solve for y.
I see BobP. set this up, but let me clarify a little.
Let x and y be the two number, as Bob had, then x+3y=600 and P=x*y. Since this is only one variable calculus, we need to express y in terms of x, so from
x+3y=600 we have y=(600-x)/3 = 200-(1/3)x
Now use that function of x for y in the product to get
P=x*[200-(1/3)x]=200x - (1/3)x^2
Find dP/dx, set to 0 and solve. You should also be in the habit of testing the endpoints and any other critical values too, sometimes these questions can throw a curve at you.
To find two numbers such that their sum plus three times the second number is 600 and their product is a maximum, we can set up the problem as follows:
Let's assume the first number is x and the second number is y. So we have the equation:
x + 3y = 600
We want to find the maximum value of the product of these two numbers, which is xy. Let's define a function U that represents the product:
U = xy
In order to find the maximum value of U, we need to take the derivative of U with respect to one of the variables (either x or y), set it equal to zero, and solve for that variable. This will give us the critical points where the maximum may occur.
Let's express y in terms of x using the first equation:
x + 3y = 600
3y = 600 - x
y = (600 - x)/3
Now substitute this value of y into our function U:
U = x * ((600 - x)/3)
U = (1/3)x * (600 - x)
U = (1/3)x * (600 - (1/3)x)
U = (1/3)x * (1800 - x)
To maximize U, we need to find the derivative of U with respect to x, set it equal to zero, and solve for x:
dU/dx = (1/3)(1800 - x) - (1/3)x
dU/dx = (1/3)(1800 - 2x)
0 = (1/3)(1800 - 2x)
0 = 1800 - 2x
2x = 1800
x = 900
So, x = 900 is a critical point. Now, we should also check the endpoints and any other critical values that may exist, as mentioned by BobP.
To find the corresponding value of y, we can substitute x = 900 into the equation y = (600 - x)/3:
y = (600 - 900)/3
y = -300/3
y = -100
Therefore, one set of numbers that satisfies the given conditions and maximizes the product is x = 900 and y = -100.
It's important to note that we should also check the other critical points and the endpoints to make sure that this is truly the maximum value.