Find the critical points of f(y):Compute the critical points of -5 y^2
To find all critical points, first compute f'(y):( d)/( dy)(-5 y^2) = -10 y:f'(y) = -10 y
Solving -10 y = 0 yields y = 0:y = 0
f'(y) exists everywhere:-10 y exists everywhere
The only critical point of -5 y^2 is at y = 0:y = 0
The domain of -5 y^2 is R:The endpoints of R are y = -∞ and ∞
Evaluate -5 y^2 at y = -∞, 0 and ∞:The open endpoints of the domain are marked in grayy | f(y)-∞ | -∞0 | 0∞ | -∞
The largest value corresponds to a global maximum, and the smallest value corresponds to a global minimum:The open endpoints of the domain are marked in grayy | f(y) | extrema type-∞ | -∞ | global min0 | 0 | global max∞ | -∞ | global min
Remove the points y = -∞ and ∞ from the tableThese cannot be global extrema, as the value of f(y) here is never achieved:y | f(y) | extrema type0 | 0 | global max
f(y) = -5 y^2 has one global maximum:Answer: f(y) has a global maximum at y = 0
Answer:
900 m
Step-by-step explanation:
l*w*h
12 m (l)*5 m (w)*15 m (h)= 900 m altogether
The system is:
i) <span>2x – 3y – 2z = 4
ii) </span><span>x + 3y + 2z = –7
</span>iii) <span>–4x – 4y – 2z = 10
the last equation can be simplified, by dividing by -2,
thus we have:
</span>i) 2x – 3y – 2z = 4
ii) x + 3y + 2z = –7
iii) 2x +2y +z = -5
The procedure to solve the system is as follows:
first use any pairs of 2 equations (for example i and ii, i and iii) and equalize them by using one of the variables:
i) 2x – 3y – 2z = 4
iii) 2x +2y +z = -5
2x can be written as 3y+2z+4 from the first equation, and -2y-z-5 from the third equation.
Equalize:
3y+2z+4=-2y-z-5, group common terms:
5y+3z=-9
similarly, using i and ii, eliminate x:
i) 2x – 3y – 2z = 4
ii) x + 3y + 2z = –7
multiply the second equation by 2:
i) 2x – 3y – 2z = 4
ii) 2x + 6y + 4z = –14
thus 2x=3y+2z+4 from i and 2x=-6y-4z-14 from ii:
3y+2z+4=-6y-4z-14
9y+6z=-18
So we get 2 equations with variables y and z:
a) 5y+3z=-9
b) 9y+6z=-18
now the aim of the method is clear: We eliminate one of the variables, creating a system of 2 linear equations with 2 variables, which we can solve by any of the standard methods.
Let's use elimination method, multiply the equation a by -2:
a) -10y-6z=18
b) 9y+6z=-18
------------------------ add the equations:
-10y+9y-6z+6z=18-18
-y=0
y=0,
thus :
9y+6z=-18
0+6z=-18
z=-3
Finally to find x, use any of the equations i, ii or iii:
<span>2x – 3y – 2z = 4
</span>
<span>2x – 3*0 – 2(-3) = 4
2x+6=4
2x=-2
x=-1
Solution: (x, y, z) = (-1, 0, -3 )
Remark: it is always a good attitude to check the answer, because often calculations mistakes can be made:
check by substituting x=-1, y=0, z=-3 in each of the 3 equations and see that for these numbers the equalities hold.</span>
Answer:
Step-by-step explanation:
Given that the figure is made up of portion of a square and a semicircle, we have;
BC ≅ AB = 6 cm
The area of semicircle BC with radius BC/2 = 3 is 1/2×π×r² = 1/2×π×3² = 4.5·π cm²
Triangle ABC = 1/2 × Area of square from which ABC is cut
The area of triangle ABC = 1/2×Base ×Height = 1/2×AB×BC = 1/2×6×6 = 18 cm²
The area of the figure = The area of semicircle BC + The area of triangle ABC
The area of the figure = 4.5·π cm² + 18 cm² = 
.
D., Because 29 x 15 is 435 but if you throw the .07 in the mix you'll have 436.05 which breaks even.
