has only one critical point at
. The function has Hessian
which is positive definite for all
, which means
attains a minimum at the critical point with a value of
.
To find the extrema (if any) along the boundary, parameterize it by
and
, with
. On the boundary, we have
Find the critical points along the boundary:
Respectively, plugging these values into
gives 11, 47, 43, and 47. We omit the first and third, as we can see the absolute extrema occur when
.
Now, solve for
for both cases:
so
has two absolute maxima at
with the same value of 47.
-1+5
--------. =
-2+9
-4
----- = 7/4
7
The formula for finding the area of a cone is pi times the radius squared times the height/3. Since we already have the base area, all we need to do is find the height over 3. The height is 6, so 6/3=2. Then 4 pi times 2=8 pi.
The answer is 8pi.
From the given volume, we know that the ratio is:
![\text {Ratio of yellow eraser to white eraser = } \sqrt[3]{ \dfrac{16}{2} }](https://tex.z-dn.net/?f=%5Ctext%20%7BRatio%20of%20yellow%20eraser%20to%20white%20eraser%20%3D%20%7D%20%20%5Csqrt%5B3%5D%7B%20%5Cdfrac%7B16%7D%7B2%7D%20%7D%20)
From the given area, we know that the ratio is:
Equate the two ratio and solve for x:
![\sqrt[3]{ \dfrac{16}{2} } = \sqrt{ \dfrac{52}{x} }](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B%20%5Cdfrac%7B16%7D%7B2%7D%20%7D%20%20%3D%20%5Csqrt%7B%20%5Cdfrac%7B52%7D%7Bx%7D%20%7D%20)
Cube both sides:
Square both sides:
Simplify each term:
Cross multiply:
Divide both sides by 256:
Cube root both sides:
<span>Ans : Note that:
sin(x) = sum(n=0 to infinity) [(-1)^n * x^(2n + 1)]/(2n + 1)!.
Then, since the series is alternating, the error in the approximation found by taking the first n terms of the series is no bigger than the n+1'th term. In other words:
E ≤ a_n+1 = x^(2n + 3)/(2n + 3)!.
(Note that a_n does not include (-1)^n, the alternating part)
We need that 1/6 ≤ x^(2n + 3)/(2n + 3)!. Given |x| < 1, n = 2 will be the least integer solution. Thus, we need 2 + 1 = 3 terms.</span>
