1-14 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} 1\ut{slug}&=14.59\ut{kg}\\ 1\ut{U.S.~bushel}&=35.238\ut{L}\\ \end{cases} $$ $$ \begin{cases} A_p\ut{\$}&=\text{Cost of pig }\ut{3.108~slug}\\ A_c\ut{\$}&=\text{Cost of corn }\ut{U.S.~bushel}\\ \cfrac{A_p}{A_c}&= 6.50\\ \end{cases} $$ $$ \begin{aligned} 3.108\ut{slug}&=3.108\ut{slug}\times\frac{14.59\ut{kg}}{1\ut{slug}}\\ &=\frac{1133643}{25000}\ut{kg}\\ \end{aligned} $$ $$ \begin{cases} B_p\ut{\$}&=\text{Cost of pig}\ut{kg}\\ B_c\ut{\$}&=\text{Cost of corn}\ut{L}\\ \cfrac{B_p}{B_c}&= ?\\ \end{cases} $$ $$ \begin{aligned} \Ans&=\cfrac{B_p}{B_c}\\ &=\cfrac{A_p\times\frac{25000}{1133643}}{A_c\times\frac{1000}{35238}}\\ &=\frac{65}{10}\cdot\frac{25000}{1133643}\cdot\frac{35238}{1000}\\ &=\frac{272675}{53983}\\ &\approx5.0511272067132245\\ &\approx5.05 \end{aligned} $$