Maxima question needs checking

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The discussion revolves around calculating the maximum area that can be enclosed by a rectangular plot of land with specific fencing costs. The farmer has a budget of £5000, with heavy-duty fencing costing £6/metre for two sides and standard fencing at £3/metre for the other two sides. The calculations involve setting up equations based on the fencing costs and using derivatives to find the maximum area. The maximum area calculated is approximately 86,803 square meters, with suggestions for using exact values and alternative methods for future calculations. The solution demonstrates a clear understanding of optimization in the context of budget constraints.
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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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