Q:

A rancher wishes to build a fence to enclose a 2250 square yard rectangular field. Along one side the fence is to be made of heavy duty material costing $16 per yard, while the remaining three sides are to be made of cheaper material costing $4 per yard. Determine the least cost, Cmin, of fencing for the rancher.

Accepted Solution

A:
Answer:The least cost of fencing for the rancher is $1200Step-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 $1200Here’s the diagram: