Sienna has 80 yards of fencing to enclose a rectangular area. Find the dimensions that maximize the enclosed area. What is the maximum area?

2 x + 2 y = 80 / : 2 2 x : 2 + 2 y : 2 = 80 : 2 x + y = 40 / − x y = 40 − x ( D x ​ : x ∈ ( 0 ; 40 ))
A re a = x y s u b s t i t u t e y = 40 − x A re a = x ( 40 − x ) = − x 2 + 40 ( i t ′ s q u a d r a t i c f u n c t i o n w h ere a = − 1 ; b = 40 ; c = 0 ) v er t e x o f p a r ab o l a : p = 2 a − b ​ → p = 2 ⋅ ( − 1 ) − 40 ​ = − 2 − 40 ​ = 20 − i t ′ s ma x x = 20 t h e n y = 40 − 20 = 20 A n s w er : d im e n s i o n s o f rec t an gu l a r i s 20 y d × 20 y d , an d a re a i s 2 0 2 = 400 y d 2 .

Answered by Anonymous | 2024-06-24

The dimensions that maximize the enclosed area with 80 yards of fencing are 20 yards by 20 yards, forming a square. The maximum enclosed area is 400 square yards. This result is derived using the relationship between perimeter and area for rectangles.
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Answered by Anonymous | 2024-09-05