Calculus Problem - Maxima, Minima?

Question

A piece of string 30 inches long has its two ends joined together and is stretched by 3 pegs so as to form a triangle. What is the largest triangular area that can be enclosed by the string?

Answer: 8(pi^0.5); I have already solved the problem assuming two sides to be equal; can some one please help me solve the problem without making that assumption - thanks

Answer 1

first let me tell u lagrange's method of undermined multipliers

let f(x ,y , z) be the function u want to find max or min value

let the constraint(connection between 3 variables) be

u(x ,y ,z) = 0

(max or min values occur at stationery values)

WORKING RULE:

1-> write F = f(x , y ,z) + k u(x ,y ,z)

2-> obtain Fx = 0 ; Fy = 0 ; Fz =0
(Fx is partial derivative of F w.r.t x)

3 -> solve the 3 eqns together with u(x ,y ,z) =0

the obtained values are statinery values


in ur problem

let x,y,z be the sides
thus the constraint is

x + y + z -30 =0

we have maximize area( or area^2)

thus
A = s(s-x)(s-y)(s-z)

here s = (x + y + z) /2 = 15

A = 15(15-x)(15-y)(15-z)

therefore

F = 15(15-x)(15-y)(15-z) + k (x+y+z-30)

diff partially w.r.t x & equating to 0

-15(15-y)(15-z) + k =0

similarly

-15(15-z)(15-x) + k =0

-15(15-x)(15-y) + k =0


from these x=y=z

thus x=y=z=10

area = sqrt (15 * 5 * 5 * 5)

Answer 2

Recall that the formula for the perimeter of a triangle is
P = a + b + c

But, we have 30 inches of string, so
30 = a + b + c

Heron's formula states that the area of any triangle is

A = sqrt[S(S - a)(S - b)(S - c)], where
S = 1/2(a + b + c)

In our case, since a + b + c = 30, then S = (1/2)(30) = 15, so

A = sqrt[15(15 - a)(15 - b)(15 - c)]

And I admittedly got stuck at this point because there's no way for me to express A as one variable. I think multivariable calculus is somehow involved, and I'm not curious about the answer myself.

Answer 3

nicely, the section enclosed is comparable to : A = b^2 + D*c^2 the place b is the size of the sq.'s area, c is the size of the triangle's area, and D is comparable to sqrt(3)/4, or, 0.433. We additionally comprehend that the twine is 10m long, so the finished perimeter is continuing. 10 = 4b + 3c we are able to sparkling up for b in terms of c, and get: b = (10-3c)/4 Substituting back into the section equation, we get: A = ((10-3c)/4)^2 + D*c^2 to locate the max and min, you're taking the by-made from the section equation and notice the place that's comparable to 0. The spinoff is: dA = (-15/4) + 2(D+9/sixteen)c = 0 c = (15/4)(a million/(2nd + 18/sixteen)) = a million.883 So all of us comprehend that c, the size of the triangle's leg, is a million.883m. Its perimeter is comparable to 5.649m, which makes the sq.'s perimeter equivalent to 4.351. this suggests that b is comparable to one fourth that perimeter, or a million.087m, and that the finished section is comparable to 2.717 m^2. via fact the 2nd derivate of A is comparable to a good consistent, all of us comprehend that that's the minimum section that would desire to be enclosed via the twine shapes. to locate the optimum, we are able to easily seem on the equations for section and notice that for a given length of twine, a sq. will continually have greater section than a triangle, so the optimum section is 6.25m^2, the element of a sq. with perimeter of 10m.

Answer 4

A=50(2^1/2)

:)