1. **Problem Statement:** A 216 m² rectangular pea patch is to be enclosed by a fence and divided into two equal parts by another fence parallel to one of the sides. We need to find the dimensions of the rectangle that minimize the total length of the fence and determine how much fence is needed. 2. **Define variables:** Let the length of the rectangle be $x$ meters and the width be $y$ meters. 3. **Given:** The area of the rectangle is $xy = 216$. 4. **Fence length:** The total fence includes the perimeter plus one fence dividing the patch into two equal parts parallel to one side. - If the dividing fence is parallel to the width $y$, then the total fence length $L = 2x + 3y$ (two lengths $x$, three widths $y$ because of the dividing fence). 5. **Express $y$ in terms of $x$ using the area constraint:** $$ y = \frac{216}{x} $$ 6. **Express total fence length $L$ as a function of $x$ only:** $$ L(x) = 2x + 3 \times \frac{216}{x} = 2x + \frac{648}{x} $$ 7. **Minimize $L(x)$:** Take the derivative and set it to zero: $$ L'(x) = 2 - \frac{648}{x^2} = 0 $$ Solve for $x$: $$ 2 = \frac{648}{x^2} \implies 2x^2 = 648 \implies x^2 = 324 \implies x = 18 $$ 8. **Find $y$:** $$ y = \frac{216}{18} = 12 $$ 9. **Check the second derivative to confirm minimum:** $$ L''(x) = \frac{1296}{x^3} $$ At $x=18$, $L''(18) > 0$, so minimum. 10. **Calculate the minimum total fence length:** $$ L = 2(18) + 3(12) = 36 + 36 = 72 $$ meters. **Final answer:** The dimensions that minimize the fence length are $18$ m by $12$ m, and the minimum total fence length needed is $72$ meters.