Problem 73 - Calculus Early Transcendentals

Question

Among all triangles with a perimeter of 9 units, find the dimensions of the triangle with the maximum area. It may be easiest to use Heron's formula, which states that the area of a triangle with side length \(a, b,\) and \(c\) is \(A=\sqrt{s(s-a)(s-b)(s-c)},\) where \(2 s\) is the perimeter of the triangle.

Step-by-Step Solution

Answer

The dimensions of the triangle with the maximum area are (9/4, 9/4, 9/2).

Step 1: Determine the semiperimeter

Given the perimeter is 9 units, we can easily find the semiperimeter (s) by dividing the perimeter by 2: \(s = \frac{9}{2} = 4.5\).

Step 2: Rewrite Heron's formula with semiperimeter

We will rewrite Heron's formula, using s: \(A=\sqrt{s(s-a)(s-b)(s-c)}\).

Step 3: Express one side in terms of the others

Since we know that the sum of the side lengths is equal to the perimeter, we can express one side in terms of the other two: \(a = 9 - (b+c)\).

Step 4: Substitute and simplify

Replace \(a\) in the area formula with the expression from Step 3 and simplify: \(A=\sqrt{s(s-(9 - (b+c)))(s-b)(s-c)}\).

Step 5: Apply the arithmetic mean-geometric mean inequality

To find the dimensions that maximize the area, we will use the arithmetic mean-geometric mean (AM-GM) inequality to relate the product of the side lengths to their arithmetic mean: \(\frac{b+c}{2} \geq \sqrt{bc}\).

Step 6: Square both sides and simplify

Square both sides of the inequality and simplify: \(\left(\frac{b+c}{2}\right)^2 \geq bc\), \(\frac{(b+c)^2}{4} \geq bc\), \((b+c)^2 \geq 4bc\).

Step 7: Use the inequality to find the optimal dimensions

Since the area is maximized when the inequality is an equality, we can set the inequality to be equal and solve for the dimensions: \((b+c)^2 = 4bc\). Substitute \(b+c=9-a\): \((9-a)^2=4(9b-9a)\). Solving for \(b\) yields \(b=a\).

Step 8: Determine the maximum area dimensions

Since \(b=a\), we can say that the side lengths are \(a, a\) and \(9-2a\). To ensure that the triangle is valid, we need the sum of the two smaller sides to be greater than the longest side: \(a+a > 9-2a\), \(2a > 9-2a\), \(4a > 9\), \(a > \frac{9}{4}\). The other side, \(b=a\), must also be greater than zero, as side lengths cannot be negative. Since the two sides are equal and they both must be greater than \(\frac{9}{4}\), we can say that the optimal dimensions for the triangle with a maximum area are \(a=b=\frac{9}{4}\) and \(c=9-2a=9-\frac{9}{2}=\frac{9}{2}\). Therefore, the dimensions of the triangle with the maximum area are \(($$\frac{9}{4}\), \(\frac{9}{4}\), \(\frac{9}{2}$$)\).