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Optimization is one of the uses of calculus in the real world. Let us assume we are a pizza parlor and wish to maximize profit. Perhaps we have a flat piece of cardboard and we need to make a box with the greatest volume. How does one go about this process?

This requires the use of maximums and minimums. We know that we find maximums and minimums via derivatives. Therefore, one can conclude that calculus will be a useful tool for maximizing or minimizing (collectively known as "optimizing") a situation.


Volume Example[edit]

A box manufacturer desires to create a box with a surface area of 100 inches squared. What is the maximum size volume that can be formed by bending this material into a box? The box is to be closed. The box is to have a square base, square top, and rectangular sides.

  • Write out known formulas and information
 A_{base} = x^2 \
 A_{side} = x \cdot h \
 A_{total} = 2x^2 + 4x \cdot h = 100
 V = l \cdot w \cdot h = x^2 \cdot h
  • Eliminate the variable h in the volume equation
 2x^2 + 4xh = 100 \
 x^2 + 2xh = 50 \
 2xh = 50 - x^2 \
 h = \frac{50 - x^2}{2x}
 V = (x^2) \left( \frac{50 - x^2}{2x} \right)

 = \frac{1}{2} (50x - x^3)

  • Find the derivative of the volume equation in order to maximize the volume
 \frac{dV}{dx} = \frac{1}{2} (50-3x^2)
  • Set  \frac{dV}{dx} = 0 and solve for  x \
 \frac{1}{2} (50-3x^2) = 0
 50-3x^2 = 0 \
 3x^2 = 50 \
 x = \pm \frac{\sqrt{50}}{\sqrt{3}}
  • Plug-in the x value into the volume equation and simplify

 V = \frac{1}{2} \left[ 50 \cdot \sqrt{\frac{50}{3}} - \left( \sqrt{\frac{50}{3}} \right) ^3 \right] =  68.04138174.. \

Answer:  V_{max} = 68.04138174.. \ 

Volume Example II[edit]

Open-top box.svg

It is desired to make an open-top box of greatest possible volume from a square piece of tin whose side is \alpha, by cutting equal squares out of the corners and then folding up the tin to form the sides. What should be the length of a side of the squares cut out?

If we call the side length of the cut out squares x, then each side of the base of the folded box is \alpha-2x, and the height is x. Therefore, the volume function is V(x)=x(\alpha-2x)^2=x(\alpha^2-4\alpha x+4x^2)=\alpha^2 x - 4\alpha x^2 + 4x^3.

We must optimize the volume by taking the derivative of the volume function and setting it equal to 0. Since it does not change, \alpha is treated as a constant, not a variable.

V(x)=\alpha^2 x - 4\alpha x^2 + 4x^3

V'(x)=\alpha^2 - 8\alpha x + 12x^2

0=12x^2 - 8\alpha x + \alpha^2

We can now use the quadratic formula to solve for x:

x=\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

x=\frac{-(-8\alpha) \pm \sqrt{(-8\alpha)^2 - 4(12)(\alpha^2)}}{2(12)}

x=\frac{8\alpha \pm \sqrt{64\alpha^2 - 48\alpha^2}}{24}

x=\frac{8\alpha \pm \sqrt{16\alpha^2}}{24}

x=\frac{8\alpha \pm 4\alpha}{24}

x=\frac{\alpha}{6}, \frac{\alpha}{2}

We reject x=\frac{\alpha}{2}, since it is a minimum (it results in the base length \alpha-2x being 0, making the volume 0). Therefore, the answer is x=\frac{\alpha}{6}.

Sales Example[edit]

Calculus Graph-Finding Maximum Profit.png

A small retailer can sell n units of a product for a revenue of r(n)=8.1n and at a cost of c(n)=n^3-7n^2+18n, with all amounts in thousands. How many units does it sell to maximize its profit?

The retailer's profit is defined by the equation p(n)=r(n) - c(n), which is the revenue generated less the cost. The question asks for the maximum amount of profit which is the maximum of the above equation. As previously discussed, the maxima and minima of a graph are found when the slope of said graph is equal to zero. To find the slope one finds the derivative of p(n). By using the subtraction rule p'(n)=r'(n) - c'(n): p(n) = r(n) - c(n)\, p'(n) = \frac{d}{dn}\left[8.1n\right]-\frac{d}{dn}\left[n^3-7n^2+18n\right]\, = -3n^2+14n-9.9\,

Therefore, when -3n^2+14n-9.9\,=0 the profit will be maximized or minimized. Use the quadratic formula to find the roots, giving {3.798,0.869}. To find which of these is the maximum and minimum the function can be tested:

p(0.869) = - 3.97321, p(3.798) = 8.58802

Because we only consider the functions for all n \ge 0 (i.e., you can't have n=-5 units), the only points that can be minima or maxima are those two listed above. To show that 3.798 is in fact a maximum (and that the function doesn't remain constant past this point) check if the sign of p'(n) changes at this point. It does, and for n greater than 3.798 P'(n) the value will remain decreasing. Finally, this shows that for this retailer selling 3.798 units would return a profit of $8,588.02.

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