posted by Samantha
4. Find the tangent line approximation, L(x), of f(x)=x^2/3 at x = 8.
L(x) = 4/3x-20/3
L(x) = 2/3x+8
L(x) = 4(x-8)
L(x) = 1/3x + 4/3
*L(x) = 4/3x - 8/3
5. A balloon is rising at a constant speed of 5 ft/sec. A boy is cycling along a straight road at a constant speed of 15 ft/sec. When he passes under the balloon, it is 5 feet above him. Approximately how fast is the distance between the boy and the balloon increasing 3 seconds after he has passed underneath it?
6. A factory is manufacturing a rectangular storage container with an open top. The volume of the container is 10 ft^3, and the length of the base is twice the width. The material for the base costs $10 per square foot, and the material for the sides costs $6 per square foot. Find the cheapest cost to make the container, given the conditions.
7. The edge of a cube was found to have a length of 50 cm with a possible error in measurement of 0.1 cm. Based on the measurement, you determine that the volume is 125,000 cm^3. Use tangent line approximation to estimate the percentage error in volume.
8. An inverted conical tank (with vertex down) is 14 feet across the top and 24 feet deep. If water is flowing in at a rate of 12 ft^3/min, find the rate of change of the depth of the water when the water is 10 feet deep.
9. For the function f(x)=Inx/x^2, find the approximate location of the critical point in the interval (0, 5).
I did #4 and #5 for you in your previous post
let the width of the base be x
let the length of the base be 2x
let the height be y
given: Volume = 10
2x^2y = 10
y = 5/x^2
which was the last choice.
cost = 10(2x^2) + 6( 2 ends + front + back)
= 20x^2 + 6(2xy + 2xy + 2xy)
= 20x^2 + 36xy
= 20x^2 + 36x(5/x^2) = 20x^2 + 180/x
d(cost)/dx = 40x - 180/x^2
= 0 for a min cost
40x = 180/x^2
x^3 = 180/40 = 9/2
x = (9/2)^(1/3)
subbing that into cost = ...
I get $163.54
Thank you so much! I'm getting number 7 as .9%, correct?
Nevermind, I figured them out! :) Thank you!