Questions: How many pounds of cashews should be mixed with the peanuts to ensure no change in the revenue? The manager buys 80 pounds of cashews and sells the mixture for 3.50 per pound. The manager should mix pounds of cashews with the peanuts.

How many pounds of cashews should be mixed with the peanuts to ensure no change in the revenue? The manager buys 80 pounds of cashews and sells the mixture for 3.50 per pound. The manager should mix pounds of cashews with the peanuts.
Transcript text: How many pounds of cashews should be mixed with the peanuts to ensure no change in the revenue? The manager buys 80 pounds of cashews and sells the mixture for $3.50 per pound. The manager should mix $\square$ pounds of cashews with the peanuts.
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Solution

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Solution Steps

Step 1: Define Variables

Let \( p_c \) be the price per pound of cashews, \( p_m \) be the price per pound of the mixture, and \( w_c \) be the weight of cashews. Given that \( p_c = 3.50 \), \( p_m = 3.50 \), and \( w_c = 80 \) pounds.

Step 2: Calculate Revenue from Cashews

The revenue from selling the cashews is calculated as: \[ R_c = w_c \cdot p_c = 80 \cdot 3.50 = 280 \]

Step 3: Set Up Revenue Equation for the Mixture

Let \( w_p \) be the weight of peanuts. The total weight of the mixture is \( w_c + w_p \). The revenue from selling the mixture is: \[ R_m = (w_c + w_p) \cdot p_m \] To ensure no change in revenue, we set \( R_m = R_c \): \[ (w_c + w_p) \cdot p_m = R_c \]

Step 4: Solve for Weight of Peanuts

Substituting the known values into the equation: \[ (80 + w_p) \cdot 3.50 = 280 \] Dividing both sides by \( 3.50 \): \[ 80 + w_p = \frac{280}{3.50} \] Calculating the right side gives: \[ 80 + w_p = 80 \] Thus, solving for \( w_p \): \[ w_p = 80 - 80 = 0 \]

Step 5: Conclusion

The calculation shows that \( w_p = 0 \), indicating that no peanuts need to be mixed with the cashews to maintain the same revenue when selling the mixture at the given price.

Final Answer

The manager should mix \(\boxed{0}\) pounds of cashews with the peanuts.

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