# Question Video: Finding the Dimensions of a Rectangle with Maximum Area given Its Perimeter Using Differentiation Mathematics • Higher Education

Find the dimensions of the rectangular corral producing the greatest enclosed area given 200 feet of fencing.

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### Video Transcript

Find the dimensions of the rectangular corral producing the greatest enclosed area given 200 feet of fencing.

In order to solve a problem like this, where we need to find a greatest or least value, we need to use optimization. The maximum and minimum values of a function will occur when the derivative dπ¦ by dπ₯ is equal to zero. If the second derivative d two π¦ by dπ₯ squared is greater than naught, we have a minimum value. If d two π¦ by dπ₯ squared is less than naught, we have a maximum value. Letβs now consider the problem in this question. Letβs consider a rectangle of width π₯ feet and length π¦ feet. The perimeter of this rectangle is equal to two π₯ plus two π¦ and the area of the rectangle is equal to π₯ multiplied by π¦.

Weβre told that we have 200 feet of fencing. Therefore, the perimeter must be equal to 200. Our first equation becomes two π₯ plus two π¦ is equal to 200. Dividing both sides of this equation by two gives us π₯ plus π¦ is equal to 100. Subtracting π₯ from both sides of the equation gives us π¦ is equal to 100 minus π₯. We can now substitute this into the second equation. The area of the rectangle is equal to π₯ multiplied by 100 minus π₯. Distributing the parentheses gives us 100π₯ minus π₯ squared. We can now begin to work out the dimensions of the greatest area by differentiating this expression.

If π΄ is equal to 100π₯ minus π₯ squared, then dπ΄ by dπ₯ is equal to 100 minus two π₯. We can differentiate each of the terms individually. Setting this equal to zero gives us 100 minus two π₯ equals zero. Adding two π₯ to both sides gives us two π₯ is equal to 100. And finally, dividing by two gives us a value of π₯ equal to 50. We know that π¦ is equal to 100 minus π₯. So we can substitute this value into the equation. 100 minus 50 is equal to 50. So π¦ is also equal to 50. Whilst it appears we have worked out the answers, it is important that we check that this value corresponds to the maximum or greatest value.

dπ΄ by dπ₯ was equal to 100 minus two π₯. We can differentiate this again to get d two π΄ by dπ₯ squared. Differentiating any constant β in this case, 100 β gives us zero. And differentiating negative two π₯ gives us negative two. Therefore, d two π΄ by dπ₯ squared is negative two. At this stage, we usually substitute in the values of π₯ into an expression. However, our answer is a constant, negative two. This is because the equation for π΄ was an n-shaped parabola and, therefore, just as one turning point. This is a maximum as d two π΄ by dπ₯ squared is negative. The dimensions that produced the greatest enclosed area are 50 feet by 50 feet.