Video: Finding the Radius of the Circular Sector with a Given Area That Has the Minimum Perimeter Using Differentiation

A sector of a circle has area 16 cmΒ². Find the radius that minimizes its perimeter, and then determine the corresponding angle πœƒ in radians.

04:47

Video Transcript

A sector of a circle has area 16 square centimetres. Find the radius that minimizes its perimeter and then determine the corresponding angle πœƒ in radians.

In order to solve a problem of this type, we need to use optimisation. The maximum and minimum values occur when d𝑦 by dπ‘₯ is equal to zero. When the second derivative d two 𝑦 by dπ‘₯ squared is greater than zero, we have a minimum value. When d two 𝑦 by dπ‘₯ squared is less than zero, we have a maximum value. As we’re dealing in radians, the area of a sector is equal to a half π‘Ÿ squared πœƒ, where πœƒ is the angle of the sector and π‘Ÿ is the radius. The perimeter of a sector is equal to π‘Ÿπœƒ, the length of the arc, plus two π‘Ÿ, two multiplied by the radius.

In this question, we’re told that the area is 16 centimetres squared. Substituting in this value gives us a half π‘Ÿ squared πœƒ is equal to 16. Multiplying both sides by two gives us π‘Ÿ squared πœƒ is equal to 32. Finally, dividing by π‘Ÿ squared gives us πœƒ is equal to 32 over π‘Ÿ squared. We can substitute this into our equation for the perimeter. The perimeter is equal to π‘Ÿ multiplied by 32 over π‘Ÿ squared plus two π‘Ÿ. In the first term, an π‘Ÿ cancels. This means that 𝑃 is equal to 32 over π‘Ÿ plus two π‘Ÿ. This can be rewritten as 32π‘Ÿ to the power of negative one plus two π‘Ÿ.

As we’re looking to minimize the perimeter, we need to differentiate this function. d𝑃 by dπ‘Ÿ is equal to negative 32π‘Ÿ to the power of negative two plus two. This can be rewritten as negative 32 over π‘Ÿ squared plus two. To find the minimum value, we then set this equal to zero. Multiplying through by π‘Ÿ squared gives us zero is equal to negative 32 plus two π‘Ÿ squared. We can then add 32 to both sides. So two π‘Ÿ squared equals 32. Dividing by two gives us π‘Ÿ squared is equal to 16. Square rooting both sides gives us π‘Ÿ is equal to positive or negative four. As we’re dealing with the length, this must be positive. Therefore, the radius is equal to four centimetres.

The corresponding angle πœƒ is therefore equal to 32 divided by four squared. Four squared is equal to 16. 32 divided by 16 is equal to two. Therefore, the angle is equal to two radians. We do need to check that this is the minimum value by working out d two 𝑦 by dπ‘₯ squared. If this value is greater than zero, then our answer is a minimum. We recall that 𝑃 was equal to 32π‘Ÿ to the power of negative one plus two π‘Ÿ. d𝑃 by dπ‘Ÿ was equal to negative 32π‘Ÿ to the power of negative two plus two.

Differentiating this again will give us d two 𝑃 by dπ‘Ÿ squared. This is equal to 64π‘Ÿ to the power of negative three as negative two multiplied by negative 32 is 64. And subtracting one from the power or exponent gives us negative three. Differentiating the constant two gives us zero. The second derivative is 64 over π‘Ÿ cubed. Substituting in our value of π‘Ÿ gives us 64 over four cubed. As four cubed is equal to 64, 64 divided by four cubed is equal to one. As this is greater than zero, our value for our radius of four centimetres is the minimum value.

The radius that minimizes the perimeter is four centimetres. And the corresponding angle is two radians.

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