The dimensions of the enclosure that is most economical to construct are; x = 14.22 ft and y = 22.5 ft
<h3>How to maximize area?</h3>
Let the length of the rectangular area be x feet
Let the width of the area = y feet
Area of the rectangle = xy square feet
Or xy = 320 square feet
y = 320/x -----(1)
Cost to fence the three sides = $6 per foot
Therefore cost to fence one length and two width of the rectangular area
= 6(x + 2y)
Similarly cost to fence the fourth side = $13 per foot
So, the cost of the remaining length = 13x
Total cost to fence = 6(x + 2y) + 13x
Cost (C) = 6(x + 2y) + 13x
C = 6x + 12y + 13x
C = 19x + 12y
From equation (1),
C = 19x + 12(320/x)
C' = 19 - 3840/x²
At C' = 0, we have;
19 - 3840/x² = 0
19 = 3840/x²
19x² = 3840
x² = 3840/19
x = √(3840/19)
x = 14.22 ft
y = 320/14.22
y = 22.5 ft
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Answer:
$31.5 i used a caculator
Step-by-step explanation:
Team 1 has -2 points and team 2 has -1 points team 2 is the winner and the results of four games would not change anything as both games 3 and 4 when added together would end up as zero extra points so it would not change anything
When lines are parallel, they are INDEPENDENT.
Parallel lines are identified as such when these lines do not intersect. They run in the same direction. They keep the same distance apart and they have the same slope.
Answer:
Option D is correct.
Step-by-step explanation:
