This might seem like a two-variable problem, but in actuality it's not - because the demand function is a sum of two components, each being independent and using only one variable, we can solve for the two separately.

Moreover, the price per unit and cost per unit is constant, so each product yields exactly $80 ($210 - $130).

Let's solve separately.

We need to maximize:

$80 * 160y/(y+4) - 1000y = $12800 * y/(y+4) - $1000*y

$80 * 170x/(x+7) - 1000x = $13600 * x/(x+7) - $1000*x

Let's go:

$12800 * y/(y+4) - $1000*y = $12800 * (1 - 4/(y+4)) - $1000y = $12800 - $51200/(y+4) - $1000*y

we analyze the derivative. $12800 is a constant, so we can skip it. Derivative of 1/(y+4) is -(y+4)^-2, derivative of $1000y is $1000.

deriv = $51200/(y+4)/(y+4) - $1000

We find the changepoints by analyzing $51200/(y+4)/(y+4) - $1000 = $0. We don't need to worry about y+4 = 0 because we cannot spend negative money on development/advertisement.

(y+4)^2 = $51.2

y+4 ~= 7.155417528 (or -y-4 = 7.1554... but it doesn't make sense because negative budget so we don't analyze).

y ~= 3.155417528

lastly, we should check that it's actually a maximum there - but it is, the original function goes to negative infinity.

rounding $1000y to the nearest dollar gives us $3155

Let's do the same for x:

$13600 * x/(x+7) - $1000*x = $13600 * (1 - 7/(x+7)) - $1000x = $13600 - $95200/(x+7) - $1000*x

deriv = $95200/(x+7)/(x+7) - $1000

$95200/(x+7)/(x+7) - $1000 = $0

(x+7)^2 = 95.2

(x+7) ~= 9.75704873412

x ~= 2.75704...