Answered

a rectangular field with a area of 8000m2 is enclosed by 400m of fencing. determine dimensions to nearest tenth of a metre



Answer :

[tex]field:x\ \times\ y\\\\ \left\{\begin{array}{ccc}x\cdot y=8000\\2x+2y=400&/:2\end{array}\right\\\left\{\begin{array}{ccc}x\cdot y=8000\\x+y=200&\to x=200-y\end{array}\right\\\\substitute\ to\ x\cdot y=8000\\\\(200-y)\cdot y=8000\\\\-y^2+200y-8000=0\\\\\Delta=b^2-4ac;\ \Delta=200^2-4\cdot(-1)\cdot(-8000)=40000-32000=8000\\\\y_1=\frac{-b-\sqrt\Delta}{2a};\ y_2=\frac{-b+\sqrt\Delta}{2a}[/tex]

[tex]\sqrt\Delta=\sqrt{8000}=\sqrt{1600\cdot5}=40\sqrt5\\\\y_1=\frac{-200-40\sqrt5}{2\cdot(-1)}=\frac{-200-40\sqrt5}{-2}=100+20\sqrt5\approx144.7\ (m)\\\\y_2=\frac{-200+40\sqrt5}{2\cdot(-1)}=\frac{-200+40\sqrt5}{-2}=100-20\sqrt5\approx55.3\ (m)\\\\x_1\approx200-144.7=55.3\ (m)\\\\x_2\approx200-55.3=144.7\ (m)\\\\Answer:Dimension of the field is:55.3m\ \times\ 144.7m[/tex]

Other Questions