Maximum and minimum problems questions with answers

Questions

1. Question 1From a given formula

   The cost, $$C$, of running a machine at speed $x$ is given by $C = 5x + \dfrac{80}{x}$, where $x > 0$. Find the minimum cost.

   Show the worked solution

   1. Step 1: Write every term as a power of x$C = 5x + 80x^{-1}$
   2. Step 2: Differentiate C with respect to x$\dfrac{dC}{dx} = 5(1) + 80(-1)x^{-2}$
   3. Step 3: For a minimum, dC/dx = 0: rearrange, then take the positive root$5 - 80x^{-2} = 0$$x^{2} = 16$$x = 4$
   4. Step 4: Check it is a minimum: put that value of x into the second derivative$\dfrac{d^{2}C}{dx^{2}} = -80(-2)x^{-3}$$= 160x^{-3}$When $x = 4$, $\dfrac{d^{2}C}{dx^{2}} = \dfrac{5}{2}$$\text{Positive, so a minimum.}$
   5. Step 5: Find the minimum value: put that value of x into C$C = 5(4) + 80(4)^{-1}$$= 40$
2. Question 2From a given formula

   The cost, $$C$, of running a machine at speed $x$ is given by $C = 2x + \dfrac{72}{x}$, where $x > 0$. Find the minimum cost.

   Show the worked solution

   1. Step 1: Write every term as a power of x$C = 2x + 72x^{-1}$
   2. Step 2: Differentiate C with respect to x$\dfrac{dC}{dx} = 2(1) + 72(-1)x^{-2}$
   3. Step 3: For a minimum, dC/dx = 0: rearrange, then take the positive root$2 - 72x^{-2} = 0$$x^{2} = 36$$x = 6$
   4. Step 4: Check it is a minimum: put that value of x into the second derivative$\dfrac{d^{2}C}{dx^{2}} = -72(-2)x^{-3}$$= 144x^{-3}$When $x = 6$, $\dfrac{d^{2}C}{dx^{2}} = \dfrac{2}{3}$$\text{Positive, so a minimum.}$
   5. Step 5: Find the minimum value: put that value of x into C$C = 2(6) + 72(6)^{-1}$$= 24$
3. Question 3Set up the formula first

   A farmer has $80$ m of fencing to make a rectangular pen against a long straight wall, so the fencing forms only three sides. The two sides perpendicular to the wall are each $x$ m long. Find the maximum area of the pen.

   Show the worked solution

   1. Step 1: Write the quantity in terms of x, then expand$\text{The side parallel to the wall} = 80 - 2x$$A = x(80 - 2x)$$A = 80x - 2x^{2}$
   2. Step 2: Differentiate A with respect to x$\dfrac{dA}{dx} = 80(1) - 2(2)x$$= 80 - 4x$
   3. Step 3: For a maximum, dA/dx = 0$80 - 4x = 0$$x = 20$
   4. Step 4: Check it is a maximum: put that value of x into the second derivative$\dfrac{d^{2}A}{dx^{2}} = -4(1)$$\text{Negative, so a maximum.}$
   5. Step 5: Find the maximum value: put that value of x into A$A = 80(20) - 2(20)^{2}$$= 800$ m²
4. Question 4Set up the formula first

   A farmer has $60$ m of fencing to make a rectangular pen against a long straight wall, so the fencing forms only three sides. The two sides perpendicular to the wall are each $x$ m long. Find the maximum area of the pen.

   Show the worked solution

   1. Step 1: Write the quantity in terms of x, then expand$\text{The side parallel to the wall} = 60 - 2x$$A = x(60 - 2x)$$A = 60x - 2x^{2}$
   2. Step 2: Differentiate A with respect to x$\dfrac{dA}{dx} = 60(1) - 2(2)x$$= 60 - 4x$
   3. Step 3: For a maximum, dA/dx = 0$60 - 4x = 0$$x = 15$
   4. Step 4: Check it is a maximum: put that value of x into the second derivative$\dfrac{d^{2}A}{dx^{2}} = -4(1)$$\text{Negative, so a maximum.}$
   5. Step 5: Find the maximum value: put that value of x into A$A = 60(15) - 2(15)^{2}$$= 450$ m²