Referencing the variables on the diagram, the 2400 m fencing restriction can be described as 2x+y=2400 or y=2400−2x Subs

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Referencing The Variables On The Diagram The 2400 M Fencing Restriction Can Be Described As 2x Y 2400 Or Y 2400 2x Subs 1 (77.8 KiB) Viewed 18 times

Referencing the variables on the diagram, the 2400 m fencing restriction can be described as 2x+y=2400 or y=2400−2x Substituting this into the standard formula, A=xy, for the the area of a rectangle gives A=xy=x(2400−2x)=2400x−2x2 So the function that must be maximized is A(x)=2400x−2x2. Differentiating shows that A′(x)=2400−4x⟹A′=0 when x=600 Since A′(x)>0 for x<600 and A′(x)<0 for x>600,A(600) is therefore the absolute maximum value of A. So the maximum area is A(600)=720,000 m2 and this is realized when the dimensions of the field are 600 m×1200 m. 1. Find the smallest possible perimeter among all rectangles with area 49 cm2. 2. Identify the largest possible product among all pairs of numbers that sum to 20 .