Suppose that the price per unit in dollars of a cell phone production is modeled by p= 65-0.0125x where xis in thousands of p represented by thousands of dollars is R=xcdot p Find the production level that will maximize revenue. square thousand phones

To find the production level that will maximize revenue, we need to find the value of x that maximizes the revenue function R.<br /><br />Given:<br />$p = 65 - 0.0125x$<br />$R = x \cdot p$<br /><br />Substituting the expression for p into the revenue function:<br />$R = x \cdot (65 - 0.0125x)$<br />$R = 65x - 0.0125x^2$<br /><br />To find the maximum revenue, we need to find the critical points of the revenue function by taking the derivative and setting it equal to zero.<br /><br />$\frac{dR}{dx} = 65 - 0.025x$<br />$65 - 0.025x = 0$<br />$x = 2600$<br /><br />Therefore, the production level that will maximize revenue is 2600 thousand phones.