Calculus Help?

Question

2. teh sum of the perimeters of an equillateral triangle and a square is 10. find the dimensions of the triangle and the square that produce a minimum total area.

3. a right circular cone is inscribed in a sphere of radius 5. determine teh dimensions of hte cone that has the maximum possible volume. hide problem

Answer 1

2. The sum of the perimeters of an equilateral triangle and a square is 10. find the dimensions of the triangle and the square that produce a minimum total area.

Let
3t = perimeter of equilateral triangle
4s = perimeter of square
A = combined area of triangle and square

Given
3t + 4s = 10
Minimize A

We have:

4s = 10 - 3t
s = (10 - 3t)/4

A = (√3/4)t² + s² = (√3/4)t² + {(10 - 3t)/4}²
A = (4√3/16)t² + (100 - 60t + 9t²)/16
A = {(9 + 4√3)t² - 60t + 100}/16

Take the derivative to find the critical values.

dA/dt = {2(9 + 4√3)t - 60}/16 = 0
2(9 + 4√3)t - 60 = 0
(9 + 4√3)t = 30
t = 30/(9 + 4√3) = 10(9 - 4√3)/11
3t = 30(9 - 4√3)/11 ≈ 5.6503548

4s = 10 - 3t = 10 - 30(9 - 4√3)/11 = 10 - (270 - 120√3)/11
4s = (110 - 270 + 120√3)/11 = (120√3 - 160)/11
4s = 40(3√3 - 4)/11 ≈ 4.3496452
s = 10(3√3 - 4)/11

A = (√3/4)t² + s²
A = (√3/4){10(9 - 4√3)/11}² + {10(3√3 - 4)/11}²
A = (100/121){(√3/4)(129 - 72√3) + (43 - 24√3)}
A = (100/121){(33√3 - 44)/4}
A = 25(3√3 - 4)/11 ≈ 2.7185282

Answer 2

in #2, I think Raj left something out. Yes, if a is the base of the triangle and b the base of the square, then 3a + 4b = 10. And again, area of the triangle will be a²/4 • √3 and area of the square will be b², and substituting (10 - 3a)/4 for b, area of the square becomes (100 - 60a + 9a²)/16, so the total area of the 2 figures becomes

(9/16)a² + (√3 / 4)a² - (60/16)a + 100/16.

To minimize this we have to find the derivative and set THAT to 0. The derivative is

(9/8)a + (√3 / 2)a - 15/4 = 0
(9/8 + √3 / 2)a = 15/4
(9 + 4√3)a = 30
a = 30 / (9 + 4√3)

Answer 3

2. 3T + 4S = 10. A = sqrt(3)/4 T*2 + S*2. Solve first relation for either T or S, substitute in the second, differentiate, equate to zero, and solve for the answer.

3. The geometry part of this is the hardest. Let r be the radius of the base of the cone, and consider a section which contains the axis of the sphere and the cone. Solve for the altitude of the cone in terms of r and the circle size, then use that to get the volume and do the usual derivative thing.

Answer 4

2.3a+4b=10
area=rt3/4 a^2+b^2
=rt3/4a^2+(10-3a/4)^2
set this to zero for maximum
=rt3/4a^2+100-15a+9a^2/16=0
=4rt3a^2-240a+1600=0
solve for a