Respuesta :
Answer:
The least cost of fencing for the rancher is $1200
Step-by-step explanation:
Let x be the width and y the length of the rectangular field.
Let C the total cost of the rectangular field.
The side made of heavy duty material of length of x costs 16 dollars a yard. The three sides not made of heavy duty material cost $4 per yard, their side lengths are x, y, y. Thus
[tex]C=4x+4y+4y+16x\\C=20x+8y[/tex]
We know that the total area of rectangular field should be 2250 square yards,
[tex]x\cdot y=2250[/tex]
We can say that [tex]y=\frac{2250}{x}[/tex]
Substituting into the total cost of the rectangular field, we get
[tex]C=20x+8(\frac{2250}{x})\\\\C=20x+\frac{18000}{x}[/tex]
We have to figure out where the function is increasing and decreasing. Differentiating,
[tex]\frac{d}{dx}C=\frac{d}{dx}\left(20x+\frac{18000}{x}\right)\\\\C'=20-\frac{18000}{x^2}[/tex]
Next, we find the critical points of the derivative
[tex]20-\frac{18000}{x^2}=0\\\\20x^2-\frac{18000}{x^2}x^2=0\cdot \:x^2\\\\20x^2-18000=0\\\\20x^2-18000+18000=0+18000\\\\20x^2=18000\\\\\frac{20x^2}{20}=\frac{18000}{20}\\\\x^2=900\\\\\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}\\\\x=\sqrt{900},\:x=-\sqrt{900}\\\\x=30,\:x=-30[/tex]
Because the length is always positive the only point we take is [tex]x=30[/tex]. We thus test the intervals [tex](0, 30)[/tex] and [tex](30, \infty)[/tex]
[tex]C'(20)=20-\frac{18000}{20^2} = -25 < 0\\\\C'(40)= 20-\frac{18000}{20^2} = 8.75 >0[/tex]
we see that total cost function is decreasing on [tex](0, 30)[/tex] and increasing on [tex](30, \infty)[/tex]. Therefore, the minimum is attained at [tex]x=30[/tex], so the minimal cost is
[tex]C(30)=20(30)+\frac{18000}{30}\\C(30)=1200[/tex]
The least cost of fencing for the rancher is $1200
Here’s the diagram:
The least cost of fencing is $ 1200.
We need to derive the formulas for the area ([tex]A[/tex]), in square yards, and the cost functions of the rectangle ([tex]C[/tex]), in monetary units:
Area
[tex]A = w\cdot l[/tex] (1)
Cost
[tex]C = (c_{w}+c_{l})\cdot w + 2\cdot c_{l}\cdot l[/tex] (2)
Where:
- [tex]w[/tex] - Width, in yards.
- [tex]l[/tex] - Length, in yards.
- [tex]c_{w}[/tex], [tex]c_{l}[/tex] - Unit costs, in monetary units per yard.
In this case, we need to determine the least possible cost by first and second derivative tests:
By (1):
[tex]w = \frac{A}{l}[/tex]
(1) in (2):
[tex]C = (c_{w}+c_{l})\cdot \left(\frac{A}{l}\right) + 2\cdot c_{l}\cdot l[/tex]
[tex]C = (c_{w}+c_{l})\cdot A\cdot l^{-1} + 2\cdot c_{l}\cdot l[/tex]
First derivative
[tex]-(c_{w}+c_{l})\cdot A \cdot l^{-2}+2\cdot c_{l} = 0[/tex]
[tex]2\cdot c_{l} = (c_{w}+c_{l})\cdot A\cdot l^{-2}[/tex]
[tex]l^{2} = \frac{(c_{w}+c_{l})\cdot A}{2\cdot c_{l}}[/tex]
[tex]l = \sqrt{\frac{(c_{w}+c_{l})\cdot A}{2\cdot c_{l}} }[/tex] (3)
Second derivative
[tex]C''= 2\cdot (c_{w}+c_{l})\cdot A\cdot l^{-3}[/tex]
[tex]C'' = 2\cdot (c_{w}+c_{l})\cdot A\cdot \left(\frac{2\cdot c_{l}}{(c_{w}+c_{l})\cdot A} \right) ^{3/2}[/tex]
[tex]C'' = 2^{5/2}\cdot \frac{c_{l}^{3/2}}{(c_{w}+c_{l})^{1/2}\cdot A^{1/2}}[/tex] (4)
The critical value leads to a minimum cost.
If we know that [tex]c_{w} = 16\,\frac{MU}{yd}[/tex], [tex]c_{l} = 4\,\frac{MU}{yd}[/tex] and [tex]A = 2250\,yd^{2}[/tex], then least cost of fencing is:
By (4):
[tex]l = \sqrt{\frac{\left(16\,\frac{MU}{yd}+4\,\frac{MU}{yd} \right)\cdot (2250\,yd^{2})}{2\cdot \left(4\,\frac{MU}{yd} \right)} }[/tex]
[tex]l = 75\,yd[/tex]
By (1):
[tex]w = \frac{2250\,yd^{2}}{75\,yd}[/tex]
[tex]w = 30\,yd[/tex]
By (2):
[tex]C = \left(4\,\frac{UM}{yd} + 16\,\frac{UM}{yd} \right)\cdot (30\,yd) + 2\cdot \left(4\,\frac{UM}{yd} \right)\cdot (75\,yd)[/tex]
[tex]C = 1200\,UM[/tex]
The least cost of fencing is $ 1200.
We kindly invite to check this question on first and second derivative tests: https://brainly.com/question/6097697
