Exercise 21 Page 646

An automatic pet feeder can be modeled by the following composite solid.

The solid consists of two smaller ones:

• a cylinder with the height of h_1= 7.5 inches, and the radius of r= 2.5 inches,
• a frustum of a cone, which is a part of a cone that lies between the base and a plane parallel to the base.Unfortunately, we do not know the value of the bottom radius of the frustum. Therefore, we have insufficient data to find the volume of the given solid. However, note that the cone with the same height of the frustum will always have smaller volume than the frustum, like below.
  
  Therefore, if a pet holder with a cone as its bottom part holds enough food for 10 days, then any other holder with a frustum as its bottom will hold enough food as well. Thus, we will find the volume of the following composite solid.
  

  The solid consists of two smaller ones:

  • a cylinder with the height of h_1= 7.5 inches, and the radius of r= 2.5 inches,
  • a cone with the height of h_2= 4 inches, and the radius of r= 2.5 inches.

  Now, let's use the formula for the volume of a cylinder, and for the volume of a cone.

  Solid	Cylinder	Cone
  Radius	r= 2.5	r= 2.5
  Height	h_1= 7.5	h_2= 4
  Volume	V_\text{Cylinder}=\pi {\color{#009600}{r}}^2{\color{#0000FF}{h_1}}	V_\text{Cone}=\dfrac{1}{3}\pi {\color{#009600}{r}}^2{\color{#0000FF}{h_2}}
  	\textcolor{darkorange}{V_\text{Cylinder}}=\pi({\color{#009600}{2.5}})^2({\color{#0000FF}{7.5}})=\textcolor{darkorange}{46.875\pi}	\textcolor{darkviolet}{V_\text{Cone}}=\dfrac{1}{3}\pi({\color{#009600}{2.5}})^2({\color{#0000FF}{4}})\approx \textcolor{darkviolet}{8.333\pi}

  Now, let's add the volumes of the cylinder and of the cone to find the volume of the food holder.
  V=\textcolor{darkorange}{V_\text{Cylinder}}+\textcolor{darkviolet}{V_\text{Cone}}

  ▼

  Substitute values and evaluate

  SubstituteII

  \textcolor{darkorange}{V_\text{Cylinder}}={\color{#0000FF}{\textcolor{darkorange}{46.875\pi}}}, \textcolor{darkviolet}{V_\text{Cone}}={\color{#009600}{\textcolor{darkviolet}{8.333\pi}}}

  V=46.875π+8.333π

  AddTerms

  Add terms

  V=55.208π

  Multiply

  Multiply

  V≈ 173.44
  Therefore, the volume of the feeder is about 173.44 cubic inches. A cat eats half a cup of food twice per day. This tells us that the cat eats one cup of food each day. One cup is about 14.4 cubic inches. Let's find how many cups of food the feeder can contain. 173.44/14.4≈ 12.04 This tells us that the feeder holds enough food for 12 days.