width of field --- x
length of field --- y , with y as the distance along the road
xy = 60000
y = 60000/x
cost = 5(2x+y) + 10y
= 10x = 15y
= 10x + 15(60000/x)
= 10x + 900000/x
d(cost)/dx = 10 - 900000/x^2 = 0 for a min of cost
10x^2 = 900000
x^2 = 90000
x = √90000 = 300
then y = 60000/300 = 200
so the field should be 200 m by 300 m , with the 200 metres along the road
minimum expense = 10x+ 15y = 3000 + 3000 = $6000
A farmer wants to fence in 60 000m^2 of land in a rectangular field along a straight road. The fencing that he plans to use along the road is $10 per meter and the fencing that he plans to use for the other three sides costs $5 per meter.
a) How much of each type of fence should he buy to keep his expanses to a minimum?
b) What is the minimum expanse?
4 answers
There were 27 students who completed a survey. There were 14 boys and 13 girls. One survey was picked at random. To the nearest hundredth, what is the probability it was completed by a girl?
13/27
read3books4minits. Idonotnoaboutfillchrips.butmabesomere
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