Lynbrook West, an apartment complex, has 100 two-bedroom units. The monthly profit (in dollars) realized from renting out x apartments is given by the following function.

p(x)=-10x^2+1700x-43,000

To maximize the monthly rental profit, how many units should be rented out?

? units

What is the maximum monthly profit realizable?

$ ?

dp/dx = 0,

or

-20x + 1700 = 0

Solve this for x for part 1, then plug this value of x into p(x) for the answer to part 2

ok I got 85 for part a and that was right. But i plugged it into part B and got 115,250....but that was wrong. What am I doing wrong?

Well, to maximize profit, we need to find the maximum point of the quadratic function p(x)=-10x^2+1700x-43,000. In this case, "x" represents the number of units rented out.

To find the number of units that should be rented out, we need to determine the x-value of the vertex of the parabola. The x-value of the vertex can be found using the formula x = -b/2a, where a, b, and c are coefficients in the quadratic equation.

For p(x) = -10x^2+1700x-43,000:
a = -10
b = 1700

Using the formula for x = -b/2a, we get:
x = -1700/(2*(-10))
x = -1700/(-20)
x = 85

So, Lynbrook West should rent out 85 units to maximize profit.

To find the maximum monthly profit realizable, we substitute the x-value obtained into the p(x) function:
p(85) = -10(85)^2+1700(85)-43,000
p(85) = -10(7225)+144,500-43,000
p(85) = -72,250+144,500-43,000
p(85) = 29,250

Therefore, the maximum monthly profit realizable is $29,250.

To find the number of units that should be rented out to maximize monthly rental profit, we need to find the vertex of the quadratic function p(x) = -10x^2 + 1700x - 43,000.

The x-coordinate of the vertex can be found using the formula x = -b / (2a), where a = -10 and b = 1700.

x = -1700 / (2(-10))
x = -1700 / (-20)
x = 85

So, to maximize profit, 85 units should be rented out.

To find the maximum monthly profit realizable, we substitute x = 85 into the function p(x):

p(85) = -10(85)^2 + 1700(85) - 43,000
p(85) = -10(7,225) + 144,500 - 43,000
p(85) = -72,250 + 144,500 - 43,000
p(85) = 29,250

Therefore, the maximum monthly profit realizable is $29,250.

To determine the number of units that should be rented out to maximize the monthly rental profit, we need to find the maximum point of the function p(x)=-10x^2+1700x-43,000.

One way to find the maximum point is by using calculus. To do this, we need to find the derivative of the function p(x) and set it equal to zero.

Let's differentiate p(x) with respect to x:

p'(x) = -20x + 1700

Now, let's find when p'(x) is equal to zero:

-20x + 1700 = 0

Solving this equation, we get:

-20x = -1700
x = 85

Therefore, to maximize the monthly rental profit, 85 units should be rented out.

To find the maximum monthly profit realizable, we need to substitute this value of x into the original function p(x).

p(85) = -10(85)^2 + 1700(85) - 43,000

Simplifying this expression, we get:

p(85) = -10(7,225) + 144,500 - 43,000
p(85) = -72,250 + 144,500 - 43,000
p(85) = 29,250

Therefore, the maximum monthly profit realizable is $29,250.