source: #youtube https://www.youtube.com/watch?v=R0juw4faGck tags: #maximum course: [[Calculus]]b status: #solved ### Problem Given $a, b$ are positive integers and $a+b=20$, find $\max \left(a^{2} b\right)$ #### Solution ##### Calculus Write $b$ in terms of $a$ $b=20-a$ Now finding max of $\left(a^2(20-a)\right)$ or $20a^2-a^3$ This is a cubic $f(a)=20a^2-a^3$ with negative end behavior, so it may have a [[Relative Maximum]] on the interval $[1,19]$. (it is certainly possible to just try all 19 values of $a$) Finding the [[Derivative of a function|derivative]] and finding its zeroes. $f^\prime(a)=40a-3a^2=0$ $a(40-3a)=0$ $a=0 \quad a=\frac{40}{3}=13.\overline3$ Trying the two nearest integers $a$ near the real location of the relative maximum $f(13)=13^2\cdot7=1183$ $f(14)=14^2\cdot6=1176$ So the maximum value is $\boxed{1183}$ ##### AM-GM Inequality Break the given sum into 3 parts, and use [[AM-GM Inequality]] $\begin{aligned}a+b=\frac{a}{2}+\frac{a}{2}+b&=20\\ \frac{a}{2}+\frac{a}{2}+b&\geq3\sqrt[3]{\frac{a^2b}{4}}\\ 20&\geq3\sqrt[3]{\frac{a^2b}{4}}\\ 8000&\geq27\left(\frac{a^2b}{4}\right)\\ 1185.\overline{185}=\frac{32000}{27}&\geq a^2b \end{aligned}$ So the geometric mean has a maximum value of 1185, occurring when $\frac{a}{2},\frac{a}{2},b$ are are equivalent. $\frac{a}{2}=b\implies a=2b$ And using this in the given sum, $a+b=2b+b=3b=20\implies b=\frac{20}{3}=7.\overline3$ So the maximum happens at $\boxed7$