# Optimization problems

## 4.7.2

- This section is like the one about Related Rates in that there's no new Math, but instead you need to construct the function yourself, an absolute extreme value of which you need to find.
- 4.7.2. Find two numbers, whose difference is 100, and whose product is a minimum.
- We need to introduce two variables for the two numbers, let it be \(x\) and \(y\).
- We have the fixed equation \(x-y=100\), and we want to find the global minimum of the function \(f(x,y)=xy\).
- We can use the fixed equation to convert the function to minimize to \(f(x)=x(x-100)=x^2-100x\).
- We have \(f'(x)=2x-100\). It has a zero at \(x=50\), we have \(f'(x)<0\) for \(x<50\), and \(f'(x)>0\) for \(x>50\).
- This means that \(f(x=50)\) is a global minimum, that is if \(x-y=100\), then to minimize the product \(xy\), we need \(x=50\) and thus \(y=-50\).

- Exercises. 4.7: 6, 8

## 4.7.14

- 4.7.14. A box with a square base and open top must have a volume of 32,000 cm\({}^3\). Find the dimensions of the box that minimzie the amount of material used.
- We have two variables: \(a\) for the length of the square base, and \(h\) for the height.
- The volume is \(a^2h=32000\).
- The function we need to maximize is \(f(a,h)=a^2+4ah\).
- From the volume, we get \(h=32000a^(-2)\), and thus \(f(a)=a^2+128000a^{-1}=a^2+2^{7}10^3a^(-1)\).
- We have \(f'(a)=2a-2^710^3a^{-2}=2(a-2^610^3a^{-2})\).
- That is, there zeroes are at \(a^3-2^610^3=a^3-(40)^3=(a-40)(a^2+40a+1600)\).
- Since \(40^2-4\cdot40^2<0\), the quadratic function \(a^2+40a+1600>0\) for all \(a\).
- Therefore, \(f'(a)\) has a zero at \(a=40\), we have \(f'(a)<0\) for \(0<a<40\), and \(f'(a)>40\).
- This shows that the minimum is at \(a=40\), and thus \(h=\frac{32000}{a^2}=\frac{2^510^3}{2^410^2}=20\).

- Exercises. 4.7: 16, 22, 30, 34, 39, 48, 54