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The Pear company sells pPhones. The cost to manufacture \( x \) pPhones is \( C(x)=-30 x^{2}+48000 x+20207 \) dollars (this includes overhead costs and production costs for each pPhone). If the company sells \( x \) pPhones for the maximum price they can fetch, the revenue function will be \( R(x)=-37 x^{2}+202000 x \) dollars. How many pPhones should the Pear company produce and sell to maximimze profit? (Remember that profiterevenue-cost.) \( x= \)

Ask by Wade Lowe. in the United States
Mar 10,2025

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Answer

The Pear company should produce and sell 11,000 pPhones to maximize profit.

Solution

We start with the profit function defined as \[ \pi(x) = R(x) - C(x) \] where \[ R(x) = -37x^2 + 202000x \] and \[ C(x) = -30x^2 + 48000x + 20207. \] Subtracting the cost from the revenue gives \[ \pi(x) = \left(-37x^2 + 202000x\right) - \left(-30x^2 + 48000x + 20207\right). \] Distribute the minus sign: \[ \pi(x) = -37x^2 + 202000x + 30x^2 - 48000x - 20207. \] Combine like terms: \[ \pi(x) = (-37x^2 + 30x^2) + (202000x - 48000x) - 20207, \] \[ \pi(x) = -7x^2 + 154000x - 20207. \] This is a quadratic function in \( x \) with the form \[ \pi(x) = ax^2 + bx + c, \] where \( a = -7 \), \( b = 154000 \), and \( c = -20207 \). Since \( a < 0 \), the parabola opens downward, meaning the maximum profit is at the vertex. The \( x \)-coordinate of the vertex is given by \[ x = -\frac{b}{2a}. \] Substitute the values of \( a \) and \( b \): \[ x = -\frac{154000}{2(-7)} = \frac{154000}{14} = 11000. \] Thus, the Pear company should produce and sell \[ x = 11000 \] pPhones to maximize profit.

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To maximize profit, we need to consider the profit function \( P(x) \), which is defined as revenue minus cost: \[ P(x) = R(x) - C(x) \] Substituting the given functions gives us: \[ P(x) = (-37x^2 + 202000x) - (-30x^2 + 48000x + 20207) \] Simplifying this leads to: \[ P(x) = -7x^2 + 153000x - 20207 \] To find the number of pPhones that maximizes profit, we can use the vertex formula for a parabola \( x = -\frac{b}{2a} \), where \( a = -7 \) and \( b = 153000 \). Calculating this: \[ x = -\frac{153000}{2 \times -7} = \frac{153000}{14} \approx 10928.57 \] Since the company can't produce a fraction of a pPhone, they should produce and sell either 10928 or 10929 pPhones. To maximize profit, rounding up is usually best, so \( x = 10929 \). --- The world of profit maximization is not just about fancy algorithms—it's about understanding the numbers that drive your business! Did you know that companies often run simulations or market tests to determine the optimal number of products to produce based on customer demand? The balance between production costs and sales price can lead to insights that go beyond mere calculations, tapping into market trends and consumer behavior! Profit maximization has real-world implications too! Think of it this way: if a company knows the exact number of products to produce, they can reduce waste, save on unnecessary overhead, and even speed up production times. Think of big names in retail who adjust their stock based on seasonal demands—it's all about maximizing that profit while keeping efficiency in check. So next time you see a sale or stock clearance, remember it’s the art of balance in action!

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