The owner of an apartment building can rent all 40 apartments if she charges \( \$ 1,700 \) per month, but she rents one fewer apartment for each \( \$ 50 \) increase in monthly (a) Write an equation that gives the revenue \( R \), from apartment rentals if she makes \( x \) increases of \( \$ 50 \) in the rent. \( R(x)= \) (b) Find the rent she should charge to maximize her revenue. \( \$ \square \) per month

To calculate the revenue \( R \) from apartment rentals, we can express the rent and the number of rented apartments based on \( x \). The rent increases by \( 50x \) dollars, so the rent per apartment becomes \( 1700 + 50x \). The number of rented apartments decreases by one for each increase in rent, resulting in \( 40 - x \) apartments rented. Therefore, the revenue function can be written as: \[ R(x) = (1700 + 50x)(40 - x) \] To find the rent that maximizes the revenue, we should determine the vertex of the quadratic function. Rewrite \( R(x) \) in standard form by expanding and simplifying: \[ R(x) = 68000 + 2000x - 50x^2 \] This is a downward-facing parabola, and the vertex (max revenue) occurs at \( x = -\frac{b}{2a} = -\frac{2000}{2(-50)} = 20 \). Thus, substituting \( x = 20 \): \[ \text{Rent} = 1700 + 50(20) = 1700 + 1000 = 2700 \] The rent she should charge to maximize revenue is \( \$ 2,700 \) per month.