The cost, in dollars, to produce \( x \) designer dog leashes is \( C(x)=9 x+6 \), and the price-demand function, in dollars per leash, is \( p(x)=105-4 x \) Find the profit function. \( P(x)= \) Find the number of leashes which need to be sold to maximize the profit. Select an answer

We start by finding the revenue function. The revenue \( R(x) \) is the product of the number of units sold and the price per unit. Given that the price-demand function is \[ p(x)=105-4x, \] the revenue function is \[ R(x)=x\cdot p(x)=x(105-4x)=105x-4x^2. \] The profit function \( P(x) \) is defined as revenue minus cost. Since the cost function is \[ C(x)=9x+6, \] the profit function becomes \[ P(x)=R(x)-C(x)= (105x-4x^2) - (9x+6). \] Simplify the expression: \[ P(x)=105x-4x^2-9x-6= -4x^2+96x-6. \] Thus, the profit function is \[ P(x)=-4x^2+96x-6. \] To find the number of leashes that need to be sold to maximize the profit, we identify the vertex of the quadratic function \( P(x) \). For a quadratic function in the form \[ P(x)=ax^2+bx+c, \] the vertex occurs at \[ x=-\frac{b}{2a}. \] Here, \( a=-4 \) and \( b=96 \). Substitute these values: \[ x=-\frac{96}{2(-4)}=-\frac{96}{-8}=12. \] Thus, the profit is maximized when \( x=12 \) dog leashes are sold. Final answers: \[ P(x) = -4x^2+96x-6, \quad x=12. \]