7 teachers because 3 x 9 = 27 and 96 - 27 = 69, 69 divided by 9 is 7 remainder something
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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Y - 3 = X + 2 ...............….... (1)
2X - 1 = Y ........................... (2)
From (2),
2X - 1 = Y
Y = 2X - 1
Sub (2) into (1),
(2X-1) - 3 = X + 2
2X - 1 - 3 = X + 2
2X - X = 2 + 1 + 3
X = 6
Sub x=6 into (1),
Y = 2(6) - 1
Y = 12 - 1
Y = 11
v is the correct answer hopefully that helped
