Optimization Math Example 2

Solution

1. 1
   Let the length be $l$ and width be $w$. Constraint: $2l + 2w = 200$, so $l + w = 100$, giving $w = 100 - l$.
2. 2
   Area to maximize: $A(l) = l \cdot w = l(100 - l) = 100l - l^2$.
3. 3
   Differentiate: $A'(l) = 100 - 2l$. Set equal to zero: $100 - 2l = 0 \Rightarrow l = 50$.
4. 4
   Check: $A''(l) = -2 < 0$, confirming a maximum.
5. 5
   Width: $w = 100 - 50 = 50$ m. Area: $50 \times 50 = 2500$ m².