There are two factories a and B. factory a is located on the bank a of a straight river bank, factory B and factory a are on both sides of the river, factory B is located at B 40km away from the river bank, and the vertical foot d from factory B to the river bank is 50km away from a. the two factories shall jointly build a water supply station C on this bank. The cost of water pipes from the water supply station to factory a and factory B is 3A yuan and 5A yuan respectively, Where is the water supply station C built to save the water pipe cost?

There are two factories a and B. factory a is located on the bank a of a straight river bank, factory B and factory a are on both sides of the river, factory B is located at B 40km away from the river bank, and the vertical foot d from factory B to the river bank is 50km away from a. the two factories shall jointly build a water supply station C on this bank. The cost of water pipes from the water supply station to factory a and factory B is 3A yuan and 5A yuan respectively, Where is the water supply station C built to save the water pipe cost?

According to the meaning of the question, the total freight can be saved only if point C is at an appropriate position on the line segment ad. set point C to be XKM away from point D, as shown in the figure, then BD = 40, AC = 50-x, ‡ BC = BD2 + CD2 = 402 + x2, and set the total water pipe cost as y yuan. From the meaning of the question, y = 3A (50-x) + 5ax2 + 402 (0 < x < 50), y '= - 3A + 5axx

A senior two liberal arts mathematics problem. (Application of derivative) A manufacturer manufactures and sells a beverage in spherical bottles. The manufacturing cost of the bottle is 0.8 π R ^ 2 points, where R is the radius of the bottle, and the unit is cm. It is known that the manufacturer can make a profit of 0.2 points for every 1ml of beverage sold, and the maximum radius of the bottle that the manufacturer can manufacture is 6cm Q: (1) what is the radius of the bottle that can maximize the profit of each bottle of beverage? (2) What is the smallest profit per bottle of beverage when the bottle radius is large? Please write down the process, thank you

Let the profit be YY = 4 π R ^ 3 / 3 * 0.2-0.8 π R ^ 2, so y '= 0.8 π R ^ 2-1.6 π R0 < R ≤ 6, let y' = 0 get r = 2, so the function y is a monotonic decreasing function at (0,2), [2,6] is a monotonic increasing function. Then the profit is the smallest when r = 2, y = - 16 π / 15. When r = 0, the profit is y = 0. When r = 6, y = 144 π / 5, so the bottle radius is 6cm

Five liberal arts math problems in senior three. Derivatives and their applications 1. If f (x) = x ^ 2 + 2x f '(1), then f' (0)=____ 2. Given the function f (x) = g (x) + x ^ 2, the tangent equation of curve g (x) at point (1, G (1)) is y = 2x + 1, then the slope of tangent of curve y = f (x) at point (1, f (1)) is_____ 3. Among all rectangles with area s (constant value), the perimeter of the rectangle with the smallest perimeter C is____ 4. The coordinates of the point closest to the fixed point P (0,2) on the curve y = - x ^ 2 + 4 (x > 0) are_____ 5. If the area of the triangle surrounded by the tangent of curve y = x ^ 3 at point (a, a ^ 3) (a ≠ 0), X axis and straight line x = a is 1 / 6, then a=_____ Please write down the process, thank you

1. Find the derivative on both sides of the equation f '(x) = 2x + 2F' (1)
When x = 1, f '(1) = 2 + 2F' (1)
So f '(1) = - 2, f' (x) = 2x - 4
So f '(0) = - 4
2.f'(x)=g'(x)+2x
Because the tangent equation of curve g (x) at point (1, G (1)) is y = 2x + 1
So G '(1) = 2
So f '(1) = g' (1) + 2x1 = 4. That is = f (x) the slope of the tangent at point (1, f (1)) is__ 4__
3. Solve by derivative method
If the length is a, the width is s / A
Perimeter y = 2 (a + S / a)
Y '= 2-2s / A ^ 2 = 0. When a = sqrt (s) is obtained, y is the smallest
And Ymin = 2 * 2sqrt (s)
4. Set the coordinates of the point a (x, - x ^ 2 + 4) and P (0,2)
So the slope of AP is (- x ^ 2 + 4-2) / (x-0) = - x + 2 / X
y'=-2x
Because the distance is the shortest, the tangent is perpendicular to the AP line, so the slope product is equal to (- 1)
So (- 2x) (- x + 2 / x) = 2x ^ 2-4 = - 1
So x = - √ 6 / 2
So (- √ 6 / 2,5 / 2)
5. Tangent k = 3x ^ 2 = 3A ^ 2
Let the distance between the focus of tangent and X axis and (a, 0) x = a ^ 3 / k = A / 3
S=a/3 × a^3 × 1/2=1/6
So a = 1 or - 1
It's hard to write!

Let a ∈ R, function f (x) = (x ^ 2-ax-a) e ^ X Find the minimum value of function f (x) on [- 2,2]

Let f '(x) = 0, the solution is x = 2 or x = a. ① a ≥ 2, then when x ∈ (2,2), f' (x) < 0, and the function f (x) monotonically decreases on (2,2). Therefore, when x = 2, the function f (x) obtains the minimum value, which is f (2) = (43A) e 2. ② 2 < a < 2

If the derivative of function f (x) is f ′ (x) and f (x) = 3x ^ 2 + 2xf ′ (2), then f ′ (5) = -------?

∵f(x)=3x^2+2xf'(2),
∴f'(x)=6x+2f'(2),
∴f'(2)=12+2f'(2),
That is, f '(2) = - 12
∴f(x)=3x^2-24x,
∴f'(x)=6x-24,
So f '(5) = 6

Point m moves in a straight line, and the distance from the starting point after T seconds is s = (1, 4) t...... 4-4t...... 3 + 16t...... 2, then the time when the speed is 0 is A4S end B8s end C0s, end of 8s D0S, 4S, 8s end Yes: S = (1 / 4) T ^ 4-4t ^ 3 + 16t ^ 2

ds/dt=t^3-12t^2+32t
=t(t-4)(t-8)
Choose D