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:
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Step-by-step explanation:
He saved 7(3), which is equal to 21 dollars saved. Hope it helps!
Given:
y = -2x + 4
2y = -6x + 12
2y = -6x + 12
2(-2x + 4) = -6x + 12
-4x + 8 = -6x + 12
-4x + 6x = 12 - 8
2x = 4
x = 4/2
x = 2
y = -2x + 4
y = -2(2) + 4
y = -4 + 4
y = 0
2y = -6x + 12 ; x = 2 ; y = 0
2(0) = -6(2) + 12
0 = -12 + 12
0 = 0
