LT 3: I can develop quadratic revenue and profit functions and determine the maximum or minimum of the quadratic function. CJ Real Estate Co. is building a new housing development. They predict the demand equation for its houses is given by \( q=-0.04 p+14,250 \), where \( q \) is the number of houses they can build if the price \( \$ p \) per house. At what price should it sell its houses to get the largest revenue?
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The Deep Dive
To maximize revenue, you first need to identify the revenue function, which is given by \( R(q) = p \cdot q \). By substituting the demand equation for \( q \) in terms of \( p \), you get \( R(p) = p \cdot (-0.04p + 14250) \), which simplifies to \( R(p) = -0.04p^2 + 14250p \). The maximum revenue occurs at the vertex of this downward-opening parabola, which can be found using the formula \( p = -\frac{b}{2a} \). Here, \( a = -0.04 \) and \( b = 14250 \), so the optimal price is \( p = \frac{14250}{0.08} = 178125 \). After calculating the price, you can easily determine the maximum revenue using this optimized price in the revenue function. It’s crucial to analyze your findings further to ensure you're maximizing revenue as gaining insight into pricing strategies can significantly impact how successful CJ Real Estate Co. might be in their new housing development. Keep in mind the pricing has to remain competitive, so it's worth exploring what similar developments are charging!