Has 320 yards of fencing to enclose a rectangular area. find the dimensions of the rectangle that maximize the enclosed area. wh
1 answer:
Let the dimensions of the rectangular area be a × b.
The perimeter of the rectangular area is 320 yards,
so
2a+2b=320
dividing by 2:
a+b=160
b=160-a
So the width and the length of the rectangular area are a and 160-a.
The area of the rectangle is a function depending on a :
we can see that A is a quadratic function, so its graph is a parabola. The parabola opens downwards because if we write A(a) in the general form -a^2+160a, we see that the coefficient of a^2 is negative.
Thus the maximal value the function can take is the y-coordinate of the vertex.
, so the roots are easily determined as 0 and 160.
the midpoint of 0 and 160 is 80.
the vertex is (80, A(80)) =(80, 80*80)=(80, 6400)
Answer: 6400 square yard.
The perimeter of the rectangular area is 320 yards,
so
2a+2b=320
dividing by 2:
a+b=160
b=160-a
So the width and the length of the rectangular area are a and 160-a.
The area of the rectangle is a function depending on a :
we can see that A is a quadratic function, so its graph is a parabola. The parabola opens downwards because if we write A(a) in the general form -a^2+160a, we see that the coefficient of a^2 is negative.
Thus the maximal value the function can take is the y-coordinate of the vertex.
the midpoint of 0 and 160 is 80.
the vertex is (80, A(80)) =(80, 80*80)=(80, 6400)
Answer: 6400 square yard.
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