Hi there

I need a little assistance on solving the following equations.



You have $ 3600 to spend on fencing for three separate rectangular pastures, all with the same dimensions. A local contracting company tells the farmer that they can build the fence for $6.25/m. What is the largest total area that you can have fenced for that price?

*This question is kind of vague, as it does not specify if the rectangular pastures are side by side... although I think it is safe to predict they are spaced out



Here's the second question... this one deals with a parabola. I have not yet learned what 'the optimum value of y' refers to.

If y=a(x-1)squared+q, what is the optimum value of y?


Thank you for your help!

For the first question :

you have 3600$ and contractor's rate is 6.25$/m , so you can have at most 576m of fence for three rectangles with same dimensions. That means you can allocate 192m of fence for each rectangle. For a fixed length of perimeter the largest rectangular area you can cover up is actually a square ( you can verify it using calculus). So your rectangles will be squares whose sides are 48m long. So the total area you can cover is 48*48*3 = 6912

For the second question :

optimum y usually means maximum or minimum possible value of y under the given circumstances. If you are allowed to use calculus then just differentiate the equation and look for a minima/maxima. Value of y at the point of maxima/minima is the answer.

Hope this helps :)

Yes, that did help! Thank you for your swiftiness, it was much appreciated :)

Your first question sounds like a calculus question. Let x and y be the dimensions of the rectangular pastures. You then need an expression for the area of the 3 pastures which you wish to maximize. You'll also need an expression for the perimeters of the pastures, which you'll set equal to 3600, then you'll solve for y in terms of x so that you can express the area in terms of just x. Then you'll calculate the derivative of your area expression to find the critical value for x. Then you'll plug that in to your y expression and the problem will be solved. As for your second question, I presume that it refers to the maximum or minimum value of y. It's then just a matter of differentiating y and the resulting expression will give you the optimum value of y.

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Originally Posted by ila20

Yes, that did help! Thank you for your swiftiness, it was much appreciated :)

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You are welcome :)