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Solve the equation x + y + (x + Y / 2) = 15 3x + 5Y + 4 (x + Y / 2) = 62
Simplification
x+y=10
5x+7y=62
5x+5y+2y=62
2y=62-50
y=6
x=4
In the equation 3x + 5Y = - 3, when 3x + y = 1, Y-X =?
In 3x + 5Y = - 3,
3X+Y=1
4Y=-4
Y=-1
X=2/3
Y-X=-5/3
Quick solution: find the maximum value of the function y = 2Sin (2x - π / 4) - 2 and the set of the corresponding values of the most independent variable x
When sin (2x - π / 4) = 1, y max = 2-2 = 0, then 2x - π / 4 = π / 2 + 2K π, so 2x = 3 π / 4 + 2K π,
So x = 3 / 8 + K, K is an integer
When sin (2x - π / 4) = - 1, y max = - 2-2 = - 4, then 2x - π / 4 = - π / 2 + 2K π, so 2x = - π / 4 + 2K π,
So x = - 1 / 8 + K, K is an integer
(you can write the set form yourself)
[calculus problem] given the function derivative and function value, find this function (details)
Find the function y = f (x). In the domain of definition (- π / 2, π / 2), the derivative is dy / DX = TaNx and satisfies the condition that f (3) = 5
F (x) = ∫ (upper x, lower 3) tank DT + 5
How is the upper and lower of ∫ determined? Why is the lower of ∫ 3? I thought the upper and lower of ∫ should be interval (here I think the upper and lower should be π / 2 and - π / 2)
∫ (upper x, lower 3) tant DT = f (x) - f (3)
F (3) = 5, so f (x) = ∫ (upper x, lower 3) tant DT + 5
Derivative of function by calculus
1. F (x) = under root (5x-3)
I change the root to (5x-3) ^ 0.5, and then I won't
2.y=5/cotx
How to find the trigonometric function?
thx!
1、
y=x^n,y'=n*x^(n-1)
So here f '(x) = 0.5 (5x-3) ^ (- 0.5) = 1 / [2 √ (5x-3)]
2、
u=cotx
y=5/u=5u^(-1)
y'=5*(-1)*u^(-2)*u'=[-5/(cot²x)]*(-csc²x)
=5*(1/sin²x)/(cos²x/sin²x)
=5sec²x
Or y = 5 TaNx
y'=5sec²x
The maximum value and derivative of function
It is known that x is greater than or equal to 0, y is greater than or equal to 0, and the maximum value of X + 3Y is ()
A36 B18 C25 D42
Sorry, if we know that x is greater than or equal to 0, y is greater than or equal to 0, x + 3Y = 9, then the maximum value of X squared by Y is ()
A36 B18 C25 D42
1/2x+1/2x+3y=9
So when 1 / 2x = 1 / 2x = 3Y = 3
1/2x*1/2x*3y
Finding the maximum value of function with derivative
In order to make a 300 m Λ 3 uncovered cylindrical reservoir, the unit cost of the bottom of the reservoir is twice that of the surrounding. How to design the size of the reservoir to make the total cost lowest?
Let R be the bottom radius, H be the height, and m be the surrounding unit cost
The total cost is f = m * 2 π RH + 2m * π R ^ 2 = 2m π (R ^ 2 + RH)
And π R ^ 2H = 300, so h = 300 / (π R ^ 2)
f=2mπ[r^2+300/(πr)]=2mπr^2+600m/r=f(r)
If this function has extremum, then f '(R) = 4m π r-600m / R ^ 2 = 0
Then 4m π r = 600m / R ^ 2, R ^ 3 = 150 / π,
Then r = (150 / π) ^ (1 / 3), H = 300 / (π R ^ 2) = 300 / [π (150 / π) ^ (2 / 3)] = 2 (150 / π) ^ (1 / 3) = 2R
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