# Question Video: Creating Linear Inequalities with One Variable Mathematics • 8th Grade

The diagram shows a cylinder of radius π and height β, and a sphere of radius π. The total surface area of the sphere is no less than that of the cylinder. Write an inequality connecting β and π. The cylinder total surface area = 2ππβ + 2ππΒ². The sphere total surface area = 4ππΒ².

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### Video Transcript

The diagram shows a cylinder of radius π and height β and a sphere of radius π. The total surface area of the sphere is no less than that of the cylinder. Write an inequality connecting β and π. The cylinder total surface area equals two ππβ plus two ππ squared. The sphere total surface area equals four ππ squared.

Letβs start by recapping that the surface area of a three-dimensional shape is the area of all its faces. For the cylinder, the two ππ squared is the area of the circle at the top and the base of the shape. And the two ππβ is the area of the rectangle that would make up the vertical section on the cylinder. For the sphere, the four ππ squared is the standard formula we would use to calculate its surface area.

So letβs start by noting down the surface areas of our cylinder and sphere. We need to think of an inequality that connects our two ππβ plus two ππ squared and the four ππ squared of the surface area of our sphere. An inequality will be a symbol such as greater than, greater than or equal to, less than, or less than or equal to.

Weβre told that the surface area of the sphere is no less than that of the cylinder. This means that the sphere can be equal or greater than the cylinder, the same size, or bigger than it. So our cylinder must be equal to or less than the sphere. So we can use the inequality less than or equal to, giving us two ππβ plus two ππ squared is less than or equal to four ππ squared.

Letβs see if we can simplify this inequality. We can start by factoring. On the left-hand side, we can see that both terms have two, a π, and an π in them. We start by writing two ππ and then our parentheses. For the first term in our parentheses, we must think, what do we multiply two ππ by to get two ππβ? And it must be β. For the second term in our parentheses, we think, what do we multiply two ππ by to get two ππ squared? And that must be π.

So weβve factored the left-hand side to give us two ππ and then in parentheses β plus π. On the right-hand side, we have four ππ squared. And we can see that we also have the similar factor of two ππ on this side. So for the missing term in our parentheses, we think, what do we multiply two ππ by to get four ππ squared? And the value would be two π.

So now in our inequality, we can divide both sides by two ππ, which would give us β plus π is less than or equal to two π. And now as we have β plus π on the left-hand side, to get β by itself, we could subtract π from both sides. So β is less than or equal to two π minus π. So β is less than or equal to π. This means that weβve written an inequality that connects β and π. So β is less than or equal to π.