Answer :
The minimum amount of fence that he can use is 1600 ft.
Given ;
Area of rectangular = 80,000 sq. ft
i.e. xy = 80,000
Total fencing needed ,
x + y + x + y + 2y = 2x + 4y
To minimize 2x + 4y, constrained to xy = 80,000
i.e. xy = 80,000
⇒ y = 80,000/x
⇒ so, s = 2x + 4 (80,000/x) to be minimized
⇒ To minimum value at ds/dx = 0
i.e. s = 2x + 4 (80,000/x)
= 2x + (3,20,00/x)
⇒ Now, ds/dx = 0 = 2 + (3,20,000) - [tex]\frac{1}{x^{2} }[/tex]
⇒ [tex]\frac{320000}{x^{2} } = 2[/tex]
⇒ [tex]x^{2} = 160000[/tex]
⇒ x = 400
∴ x = 400 ft
y = 80,000/400 = 200
∴ y = 200 ft
Total frencing is
⇒ 2x + 4y = 2 × 400 + 4 × 200
= 800 + 800
= 1600 ft
∴ The minimum amount of fence that he can use is 1600 ft.
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