Problem

Solution to Problem 1:

• This problem has been solved graphically. Here we solve it more rigorously using the first derivative.
  
• We first use the formula of the volume of a pyramid to write the equation:
  (1 / 3) h x ² = 1000
  
• We now use the the formula of the surface area of a pyramid to write a formula for the surface area S of the given pyramid. In this problem we have a square pyramid, hence:
  S = x * √ [ h ² + (x/2) ² ] + x * √ [ h ² + (x/2) ² ] + x * x
  = 2 x √ [ h ² + (x/2) ² ] + x ²
  
• Solve the equation (1 / 3) h x ² = 1000 for h to obtain:
  h = 3000 / x ²
  
• Substitute h in the surface area formula by 3000 / x ² to obtain a formula for S in terms of x (x positive) only and rewrite it as follows:
  S = √ [ (36 10 ⁶ + x ⁶) ] / x + x ²
  
• Let constant k = 36 10 ⁶ and differentiate S with respect to x.
  dS / dx = [ 3 x ⁶ (k + x ⁶) ^-1/2 - (k + x ⁶) ^1/2 ] / x ² + 2 x
  
• Multiply numerator and denominator by (k + x ⁶) ^1/2 and simplify.
  dS / dx = [ 2 x ⁶ - k ] / [ x ² (k + x ⁶) ^1/2 ] + 2x
  
• A graph of dS / dx is shown below. For x > 0, dS / dx has a zero and is negative to the left of that zero and positive to the right of the zero. This means that S has a minimum value that may be located by setting dS / dx = 0 and solve for x.
  
  [ 2 x ⁶ - k ] / [ x ² (k + x ⁶) ^1/2 ] + 2 x = 0
  
• Rewrite as.
  [ 2 x ⁶ - k ] = - 2 x ³ (k + x ⁶) ^1/2
  
• Let u = x ³ and u ² = x ⁶ and rewrite the above as follows:
  [ 2 u ² - k ] = - 2 u (k + u ²) ^1/2
  
• Square both sides:
  4 u ⁴ + k ² - 4k u ² = 4 u ² (k + u ²)
  
• Simplify to obtain:
  k ² - 4k u ² = 4 k u ² k
  
• Solve for u: (u positive since x positive)
  u = √ [ k / 8 ]
  
• Finally solve for x to obtain:
  x = [ √ [ k / 8 ] ] ^1/3
  
• Substitute k by its value and calculate x:
  x = 12.8 cm (rounded to 1 decimal place).
  
• Below are shown the graphs of the surface area and its derivative.
  

More references on calculus problems