Question

what is the max area of an isosceles triangle with two side lengths of 5? Use calculus optimization of max and min please.

Answer 1

\[\sqrt{3}\times6.25\]

Answer 2

how did you figure that out?

Answer 3

u know the fact area of an eqilateral trngle s most

Answer 4

its not equilateral though?

Answer 5

or u can use max. or min. concept

Answer 6

y not

Answer 7

yeah, but how do you use the max or min concept??

Answer 8

its says its an isosceles.

Answer 9

take the base x
and figure rhe area with the help of x
then difrntiate it

Answer 10

eqilateral is also isosceles

Answer 11

but is the equilateral area the MAXIMUM area it could be? thats the thing.. and i can take the base and find the area and differentiate..but i can't get it to come out to an answer...

Answer 12

u can use area = 0.5 *5*5 * sin x where x = vertex angle (if i remember the formula crrectly)

Answer 13

okay this is what i have
area= .5(x)((25-((x^2)/4))^(1/2))
when i try to differntiate that is when i get stuck

Answer 14

use pythagorus

Answer 15

i dont know its a right triangle! i cant use pythag.

Answer 16

den do diffrntiate
and see it comes
that

Answer 17

the diffientiation is what im getting stuck at.

Answer 18

join the vertex to mid pnt of base
then apply pyth.

Answer 19

okayy. that is not getting me the max. i have to use CALCULUS.

Answer 20

who restrict u

Answer 21

my teacher..

Answer 22

A = (25-x^2)^(1/2) * x where x = half base
dA/dx = (25-x^2)^(1/2) + x* (1/2)* (25-x^2)^(-1/2)* -2x [by product rule)
dA/dx = (25-x^2)^(1/2) - x^2 / (25-x^2)^(1/2)
= 25 - x^2 - x^2 = 0 for max/min
x = sqrt 12.5

max area is when length of base = 2* sqrt 12.5
max area = height * sqrt 12.5
= (25-12.5)^(1/2) * sqrt 12.5