Geometrical Applications of Calculus (2 Viewers)

[namburger]

Ugh... MORE help needed;

Find the greatest and least values of the function f(x) = 4x³ - 3x² - 18x in the domain -2 ≤ x ≤ 3.

Man I suck...

f(x) = 4x³ - 3x² - 18x
f'(x) = 12x^2 - 6x - 18
let f'(x) = 0
2x^2 - x - 3 = 0
(2x-3)(x+1) = 0

Therefore stationary pts occur at x=-1, 3/2
At x = -1, y = 11
At x = 3/2, y = -20.25

Max at x=-1, Min at x=3/2

EDIT: actually it asks for greatest and least value in the domain, so therefore you must also test x=-2 and 3
at x=-2, y=-8
at x = 3, y=27

THEREFORE Maximum Y value is 27, Minimum Y value is -20.25

 

[cwag]

Ugh... MORE help needed;

Find the greatest and least values of the function f(x) = 4x³ - 3x² - 18x in the domain -2 ≤ x ≤ 3.

Man I suck...

for greatest and least values we must find turning points within that domain, and check if they are bigger than the endpoints in this case -2 and 3

therefore f(x) = 4x³-3x<sup>2/sup]-18x
f'(x)=12x2-6x-18
f(x)= 0 when 4x2-2x-6=0
(4x-6)(x+1)=1
x= 3/2, or -1

when x = 3/2, f(x)= -20.25
when x = -1, f(x)=11
when x = -2, f(x) = -8
when x = 3, f(x) = 27

therefore the greatest value is 27 (at x=3)
the smallest value is -20.25 (at x=3/2, turning point)</sup>

 

[FDownes]

What the heck? Those are exactly the same x-values I got the first I ran through the question, but I didn't get the same y-values... I must have substituted the numbers in to the derivative or something... Gah! I'm such an idiot.

Thanks guys.

 

[FDownes]

Here we go again;

Two circles have radii r such that r + s = 25. Show that the sum of the areas of the circles is least when r = s.

 

[namburger]

Here we go again;

Two circles have radii r such that r + s = 25. Show that the sum of the areas of the circles is least when r = s.

Area of first circle = pi r^2
Area of second circle = pi (25-r)^2

A = pi[r^2 + 625 - 50r + r^2]
= pi [2r^2 -50r +625]

A' = pi[4r-50]
let A' = 0
r= 12.5
if r =12.5, therefore s =12.5