Answer
a) $A=x(\frac{100-x}{2})$
b) (graph is shown in the image below)
let the y-axis becomes the Area and the x-axis as the length in x.
Based on the graph shown, the length of 50 meters will yield the maximum amount of area for the pasture.
c) dimensions: 25x50 m
Work Step by Step
a)
$2y+x=100$
$2y=100-x$
$y=\frac{100-x}{2}$
$A=xy$
$A=x(\frac{100-x}{2})$
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b)
(graph is shown in the image below)
let the y-axis becomes the Area and the x-axis as the length in x.
Based on the graph shown, the length of 50 meters will yield the maximum amount of area for the pasture.
------------------------------
c)
$A=x(\frac{100-x}{2})$
$=-x^2+50x$
$A=-\frac{1}{2}(x^2-100x+2500)+1250$
$A=-\frac{1}{2}(x-50)^2+1250$
$x=5 m$
$y=\frac{100-50}{2}$
$=25 m$
dimensions: 25x50 m
