Among ail right circular cones with a slant height of th, what are the dimensions (radius and heght) that maximize the volume of the cone9 the siant height of a cone in the distane frant edge of the base to the vertex Let V be the volume of the cone What is the objective function in terms of the heigh of the cone, th? \[ V=\frac{\pi}{3}\left(225 h-h^{3}\right) \] (Type an expression Type an exact answer, using \( n \) as needed) The interval of interest of the objective function is \( [0,15] \) (Simplify your answer. Type your answer in interval notation) To maximize the volume of a right circular cone with a slant height of t5, the height must be (Type exact answers, using radicals as needed) \( \square \) and the radius of the base must be \( \square \) I

To maximize the volume of a right circular cone with a fixed slant height of \( 15 \), we need to determine the optimal height \( h \) and radius \( r \) of the cone. Here's a step-by-step breakdown: ### 1. **Objective Function** The volume \( V \) of a cone is given by: \[ V = \frac{1}{3} \pi r^2 h \] Given the slant height \( l = 15 \), we relate \( r \) and \( h \) using the Pythagorean theorem: \[ r^2 + h^2 = l^2 \implies r = \sqrt{225 - h^2} \] Substituting \( r \) into the volume formula: \[ V(h) = \frac{1}{3} \pi (225 - h^2) h = \frac{\pi}{3} (225h - h^3) \] **Objective Function:** \[ V(h) = \frac{\pi}{3} (225h - h^3) \] ### 2. **Interval of Interest** Since the height \( h \) must be positive and cannot exceed the slant height: \[ h \in [0, 15] \] ### 3. **Maximizing the Volume** To find the height \( h \) that maximizes the volume, take the derivative of \( V(h) \) with respect to \( h \) and set it to zero: \[ V'(h) = \frac{\pi}{3} (225 - 3h^2) = 0 \implies 225 - 3h^2 = 0 \implies h^2 = 75 \implies h = 5\sqrt{3} \] **Optimal Height:** \[ h = 5\sqrt{3} \] ### 4. **Determining the Radius** Using the relationship \( r^2 + h^2 = 225 \): \[ r^2 = 225 - (5\sqrt{3})^2 = 225 - 75 = 150 \implies r = 5\sqrt{6} \] **Optimal Radius:** \[ r = 5\sqrt{6} \] ### **Final Answer** To maximize the volume of a right circular cone with a slant height of \( 15 \): - **Height:** \( h = 5\sqrt{3} \) - **Radius:** \( r = 5\sqrt{6} \) \[ \boxed{ \begin{aligned} &\text{Objective Function: } V(h) = \frac{\pi}{3} (225h - h^3) \\ &\text{Interval: } [0, 15] \\ &\text{Optimal Dimensions: } \\ &\quad \text{Height } h = 5\sqrt{3} \\ &\quad \text{Radius } r = 5\sqrt{6} \end{aligned} } \]