1. At midnight, you are 60 feet due north of Jonas. You run due east at ten feet per second. Jonas walks two feet per second due north. How many seconds after midnight is Jonas closest to you? What is the corresponding minimum distance separating you and Jonas?
2. We need to build a building in the shape of a rectangular box with a capacity of 500,000 cubic feet. We require the length of the floor to be twice the width. The floor will cost $3 per square foot, the vertical walls will cost $4 per square foot, and the ceiling will cost $5 per square foot. To the nearest hundredth of a foot, what are the dimensions of the cheapest building that we can build? What is that building’s cost to the nearest dollar
the distance z between you and Jonas after t seconds is
z^2 = (60-2t)^2 + (10t)^2
For the box, we know
xyz = 500000
since y=2x,
2x^2z = 500000
x^2z = 250000
So, z = 250000/x^2
c(x) = (x)(2x)(3) + (6x)(250000/x^2)(4) + (x)(2x)(5)
= 16x^2 + 6000000/x
Now just do your minimum stuff.
1. To determine the time when Jonas is closest to you, we can break down the problem into different time intervals and calculate the minimum distance during each interval.
a) At midnight, you are 60 feet due north of Jonas. Let's assume the initial position of Jonas is the origin (0,0) on a coordinate plane.
b) You run due east at ten feet per second. This means your position can be determined by the equation x = 10t, where x is the horizontal distance traveled and t is the time in seconds.
c) Jonas walks two feet per second due north. This means Jonas' position can be determined by the equation y = 2t, where y is the vertical distance traveled and t is the time in seconds.
d) Now, we need to find the time when Jonas is closest to you. To do this, we need to calculate the distance between your position and Jonas' position at different times. The distance formula is given by D = sqrt((x2 - x1)^2 + (y2 - y1)^2).
e) In this case, x1 = 0 (initial x-position of Jonas), y1 = 0 (initial y-position of Jonas), x2 = 10t (your x-position), and y2 = 60 - 2t (your y-position). Plugging in these values into the distance formula, we get D = sqrt((10t - 0)^2 + (60 - 2t - 0)^2).
f) Simplifying the distance formula, we have D = sqrt(100t^2 + 3600 - 240t + 4t^2).
g) To find the minimum distance, we can take the derivative of D with respect to t, set it equal to zero, and solve for t.
h) Once we find the value of t, we can substitute it back into the equation for D to find the minimum distance and determine how many seconds after midnight Jonas is closest to you.
2. To find the dimensions of the cheapest building, we need to consider the cost of each component and minimize the total cost.
a) Let's assume the width of the floor is w and the length is 2w (twice the width).
b) The volume of a rectangular box is given by V = length × width × height.
c) In this case, we know the volume (V) is 500,000 cubic feet. So, we have 500,000 = 2w × w × h, where h is the height of the box.
d) Rearranging the equation, we have h = 500,000 / (2w^2).
e) The cost of the floor is given by the area of the floor (length × width) multiplied by the cost per square foot (which is $3). So, the cost of the floor is C_floor = 3 × (2w × w).
f) The cost of the vertical walls is given by the area of the four walls (2 × length × height + 2 × width × height) multiplied by the cost per square foot (which is $4). So, the cost of the vertical walls is C_walls = 4 × (2(2w × h) + 2(w × h)).
g) The cost of the ceiling is given by the area of the ceiling (length × width) multiplied by the cost per square foot (which is $5). So, the cost of the ceiling is C_ceiling = 5 × (2w × w).
h) The total cost is given by adding the cost of the floor, walls, and ceiling: C_total = C_floor + C_walls + C_ceiling.
i) Substitute the expressions for C_floor, C_walls, and C_ceiling into the equation for C_total.
j) To minimize the cost, we can differentiate C_total with respect to w, set it equal to zero, and solve for w.
k) Once we find the value of w, we can calculate the corresponding length (2w) and height (500,000 / (2w^2)) to determine the dimensions of the cheapest building.
l) Finally, substitute the dimensions into the cost equation C_total to find the cost of the building to the nearest dollar.