stevetron
  • stevetron
Suppose you have 54 feet of fencing to enclose a rectangular dog pen. The function A=27x-x^2, where x = width, gives you the area of the dog pen in square feet. What width gives you the maximum area? What is the maximum area? Round to the nearest tenth as necessary. a.width = 13.5 ft; area = 546.8 ft2 b.width = 27 ft; area = 182.3 ft2 c.width = 13.5 ft; area = 182.3 ft2 d.width = 27 ft; area = 391.5 ft2
Mathematics
chestercat
  • chestercat
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campbell_st
  • campbell_st
is this a calculus or algebra question..?
stevetron
  • stevetron
algebra
campbell_st
  • campbell_st
ok... find the line of symmetry, use \[x = \frac{-b}{2 \times a}\] so in your question b = 27 and a = -1 substitute them to find the width for the max area

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campbell_st
  • campbell_st
what you have found is the line of symmetry for the parabola. The maximum area lies on the line of symmetry so given a quadratic \[ax^2 + bx + c \] the line of symmetry is \[x = \frac{-b}{2a}\] then to find the max area, substitute the value into the original equation
stevetron
  • stevetron
so a=27*13.5-13.5^2?
campbell_st
  • campbell_st
that's correct
stevetron
  • stevetron
thank you for explaining it
campbell_st
  • campbell_st
glad to help

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