SOLUTION: Pamela has 50 feet of fencing tp put around a rectangular dog pen. Her house is used as one side of the pen. A 2 feet opening is left on another side for a gate. What length and w

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: Pamela has 50 feet of fencing tp put around a rectangular dog pen. Her house is used as one side of the pen. A 2 feet opening is left on another side for a gate. What length and w      Log On


   

Question 74829: Pamela has 50 feet of fencing tp put around a rectangular dog pen. Her house is used as one side of the pen. A 2 feet opening is left on another side for a gate. What length and width will provide a maximim area for her dog?
Answer by checkley75(3666) About Me  (Show Source):
You can put this solution on YOUR website!
FOR A RECTANCLE TO ENCLOSE THE MAXIMUM AREA IT SHOULD BE A SQUARE. THUS THREE SIDES OF THE SQUARE HERE WOULD TOTAL 50+2=52 FEET (FENCE + GATE)
SEEING AS THERE ARE ONLY THREE SIDES, THE HOUSE BEING THE FOURTH SIDE. WE DIVIDE 52 BY 3=17 1/3 FEET PER SIDE (LENGTH & WIDTH).
17.333*17.333=300.444 SQUARE FEET IS THE MAXIMUM FOR THE AREA ENCLOSED.