For symmetry reasons, the optimum must have a square base. Let the height of the walls be x and the side length be y
Volume = x^2*y = 50
Area = x^2 +4xy = x^2 + 4x*50/x^2
= x^2 + 200/x
dA/dx = 0 = 2x -200/x^2
x^3 = 100
x = 4.64 ft
y = 2.32 ft
x = length
y = width
h = height
x y h = 50
area bottom = x y
area sides = 2 x h + 2 y h
total area a = x y + 2 x h + 2 y h
h = 50/xy
a = x y + 100/y + 100/x
da/dx=x dy/dx+y-100 dy/dx /y^2-100/x^2
for min = 0
dy/dx(x-100/y^2) + y-100/x^2 = 0
x y^2 = 100 and y x^2 = 100
y (100^2/y^4) = 100
y^3 = 100
y = 4.64
x = 100/y^2 = 4.64
h = 2.32
Answer this Question
Calculus-Applied Optimization Problem - If a total of 1900 square centimeters of...
calculus optimization problem - by cutting away identical squares from each ...
Calculus - Given: available material= 1200cm^2 Box w/ square base & open top ...
Calulus/Optimization - I don't understand how to solve optimization problems, (...
Quick math help - In which step below does a mistake first appear in simplifying...
Calculus (Optimization) - A rectangular piece of cardboard, 8 inches by 14 ...
algebra - would you help me write a logic application for andy has the cards1,5...
Algebra II - Solving the formula for the indicated variable. I=prt,for r I need ...
math volume - a rectangular sheet of cardboard size 5' by 8' is to be used to ...
Pre-Calculus - I'm having a had time with this problem. Could you please help ...