Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter P - Problem Solving - Page 39: 6


a) $=\frac{3x(100-x)}{2}$ b) 37.5 x 2 = 70 c) $50\times70$ =$3750ft^2$

Work Step by Step

a) $4y+3x=300$ $4y=300-3x$ $4y=3(100-x)$ $y=\frac{3}{4}(100-x)$ $A=2xy$ $=(2x)(\frac{3}{4}(100-x)$ $=\frac{3x(100-x)}{2}$ ------------------------------------- b) the graph is shown in the image. The y-axis represents area A and the x-axis represents the length in x. By referring to the graph, the length x of 50 ft, which results in the y value as 37.5 ft will yield the maximum area of the pastures of 3750 $ft^2$. (double the y value, since the length is constituted of 2y. Thus, 37.5 x 2 = 70) ------------------------------------- c) $A=\frac{3x(100-x)}{2}$ $=-\frac{3x^2}{2}+\frac{300}{2}x$ $=-\frac{3}{2}(x^2-100x+2500)+3750$ $=-\frac{3}{2}(x-50)^2+3750$ $x=50ft$ $y=\frac{3}{4}(100-x)$ $=\frac{3}{4}(50)$ $y=37.5ft$ A=2xy =$(50)(2\times37.5)$ =$50\times70$ =$3750ft^2$ Therefore, the dimension 50 x 70 ft will yield the maximum area of $3750ft^2$
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