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Can u explain to me how to do # 8 of the practice final u gave us. I have a general idea of how to do these questions.....just can't figure this one out.

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3x+y=20 (this is what the second sentence told us). The area is A=xy. Since y=20-3x, we have that A=x(20-3x)=20x-3x^{2}. We want to maximize A. First, critical numbers: A'=20-6x and A'=0 yields x=20/6. Since A''=-6 < 0 (by the second derivative test) we have a local maximum at x=20/6. Since this is the only critical number (yielding the only peak of the function), it must be the absolute maxiumum. So, we have the absolute maximum at x=20/6. Since y=20-3x, we have that y=20-3(20/6)=10. So, to summarize, x=20/6. y=10 give the dimensions of the pens with maximal area.

Now the cost to build the fence is C=30y+35(3x) [30 per meter for the y -meters, and 35 per meter for the 3x maters of fencing]. So, the cost to build the fence with maximal area is (30)(10)+35(3)(20/6)=650 $.

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Could you show 4 and 5 please?