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SUMMARY

The discussion focuses on optimizing the area of a rectangular plot of land using two types of fencing: heavy-duty fencing at £6/metre and standard fencing at £3/metre. The farmer has a budget of £5000, leading to the equation 5000 = 12y + 6x, where y represents the heavy-duty fencing sides and x represents the standard fencing sides. The maximum area that can be enclosed is calculated to be 86803 m² by substituting values into the area formula A = x * y and using calculus to find the optimal dimensions. For improved accuracy, using exact values instead of approximations is recommended.

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a rectangular plot of land requires fencing on all 4 sides. two opposite sides will use heavy duty fencing at £6/metre, whilst the other tow sides will use standard fencing at £3/metre. if the farmer buying the fencing has £5000 to spend, what is the maximum area he can enclose?

attempt

let y = heavy duty fencing sides and x= standard fencing sides.

A=x*y
5000=6y+6y+3x+3x
5000=12y+6x...solve for x (or y)...
5000-12y=6x (divide by 6)...
833.33-2y=x...Now we will substitute this into A=x*y

A=(833.33-2y)*y
A=833.33y-2y2 ...now take derivative and set equal to 0
A'=833.33-4y=0
or...4y=833.33
y=208.33 meters.find x
5000-(12*208.33)=6x
2500=6x
x=416.66 meters

So max area is x*y or (208.33*416.66)= 86803m2.
 
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That is the correct method and solution. It may be slightly better to use the exact values, eg 2500/3, rather than 833.33, and also instead of differentiating and solving, you could use the formula for the axis of symmetry of a parabola x=-b/2a, might be quicker next time.
 

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