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?
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
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
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To answer these questions, we can use the concepts of optimization and linear equations. Let's break down the problem step by step.
a) How much of each type of fence should he buy to keep his expenses to a minimum?
To minimize expenses, we need to minimize the cost of the fence. Let's assume the length of the rectangular field along the road is x meters, and the width is y meters.
The given information tells us that the total area of the field is 60,000 square meters, so we can write the equation:
x * y = 60,000 ---(Equation 1)
The cost of the fence along the road is $10 per meter, and since there are two lengths and one width along the road, the cost of the fence along the road can be expressed as:
2x * 10 ---(Equation 2)
The cost of the fence on the other three sides is $5 per meter, and since there are two widths and one length on these sides, the cost of the fence on the other three sides can be expressed as:
2y * 5 ---(Equation 3)
To minimize the cost, we need to minimize the sum of the two costs. So, we can write the total cost equation:
Total cost = 2x * 10 + 2y * 5 ---(Equation 4)
To find the values of x and y that minimize the cost, we need to solve this system of equations simultaneously. Substituting Equation 1 into Equation 4, we get:
Total cost = 2(60,000/y) * 10 + 2y * 5
Now, we have an equation with only one variable (y). To minimize the total cost, we can take the derivative of the equation with respect to y, set it equal to zero, and solve for y.
b) What is the minimum cost?
Once we find the value of y that minimizes the cost, we can substitute it back into Equation 1 to find the corresponding value of x. Then, we can substitute the values of x and y into Equation 4 to calculate the minimum total cost.
Please note that finding the derivative and solving the resulting equation requires advanced calculus and algebraic skills. It is recommended to use appropriate software or consult a mathematics expert for assistance.