Maxima and Minima - Solved Examples

1. Find the greatest value of a2.b3.c4$a^2.b^3.c^4$ subject to the condition a+b+c=18
Sol:  Though sum of the variables are constant in this question, directly we cannot apply the rules learned above.  We have to modify the given expression to suit the above rules
Let Z = a2.b3.c4$a^2.b^3.c^4$
Z = 22.33.44.(a2)2.(b3)3.(c4)4$2^2{.3}^3{.4}^4.{\left({\displaystyle \frac{a}{2}}\right)}^2.{\left({\displaystyle \frac{b}{3}}\right)}^3.{\left({\displaystyle \frac{c}{4}}\right)}^4$
[ any question of this type, we modify ap$a^p$ as (ap)p${\left(\frac{a}{p}\right)}^p$ so on and multiply with suitable powers to make it equal to original equation]
Z will have the maximum when (a2)2.(b3)3.(c4)4${\left({\displaystyle \frac{a}{2}}\right)}^2.{\left({\displaystyle \frac{b}{3}}\right)}^3.{\left({\displaystyle \frac{c}{4}}\right)}^4$ is maximum.

But (a2)2.(b3)3.(c4)4${\left({\displaystyle \frac{a}{2}}\right)}^2.{\left({\displaystyle \frac{b}{3}}\right)}^3.{\left({\displaystyle \frac{c}{4}}\right)}^4$ is a product of 2+3+4=9 factors whose sum = 2(a2)+3(b3)+4(c4)$2\left({\displaystyle \frac{a}{2}}\right)+3\left({\displaystyle \frac{b}{3}}\right)+4\left({\displaystyle \frac{c}{4}}\right)$ = a+ b + c = 18
(a2)2.(b3)3.(c4)4${\left({\displaystyle \frac{a}{2}}\right)}^2.{\left({\displaystyle \frac{b}{3}}\right)}^3.{\left({\displaystyle \frac{c}{4}}\right)}^4$ will be maximum if all the factors are equal. i.e., if a2=b3=c4=a+b+c9=189=2${\displaystyle \frac{a}{2}=\frac{b}{3}={\displaystyle \frac{c}{4}=\frac{a+b+c}{9}=\frac{18}{9}=2}}$
So maximum value of Z = 22.33.44.(2)2.(2)3.(2)4=219.33$2^2{.3}^3{.4}^4.(2{)}^2.(2{)}^3.(2{)}^4=2^{19}{.3}^3$

Alternate method:

The greatest value of am.bn.cp$a^m.b^n.c^p$, when m, n, p being +ve integers, a+b+c is constant is given by
mm.nn.pp......(a+b+c+...m+n+p+...)m+n+p+..$m^m.n^n.p^p......{\left({\displaystyle \frac{a+b+c+...}{m+n+p+...}}\right)}^{m+n+p+..}$

By applying above concept: 22.33.44.(189)9=219.33$2^2{.3}^3{.4}^4.{\left({\displaystyle \frac{18}{9}}\right)}^9=2^{19}{.3}^3$

2. If 2x+3y=7; find the greatest value of x3.y4$x^3.y^4$
Solution: Let Z = x3.y4$x^3.y^4$
[ we change the original function by taking (2x3)3${\left({\displaystyle \frac{2x}{3}}\right)}^3$ instead x3$x^3$ and (3y4)4${\left({\displaystyle \frac{3y}{4}}\right)}^4$ instead of y4$y^4$]
 So Z = x3.y4$x^3.y^4$ = (32)3(43)4(2x3)3(3y4)4${\left({\displaystyle \frac{3}{2}}\right)}^3{\left({\displaystyle \frac{4}{3}}\right)}^4{\left({\displaystyle \frac{2x}{3}}\right)}^3{\left({\displaystyle \frac{3y}{4}}\right)}^4$

But (2x3)3(3y4)4${\left({\displaystyle \frac{2x}{3}}\right)}^3{\left({\displaystyle \frac{3y}{4}}\right)}^4$ is a product of 3 + 4 = 7 factors, whose sum = 3(2x3)+4(3y4)$3\left({\displaystyle \frac{2x}{3}}\right)+4\left({\displaystyle \frac{3y}{4}}\right)$ = 2x + 3y = 7

Therefore; (2x3)3(3y4)4${\left({\displaystyle \frac{2x}{3}}\right)}^3{\left({\displaystyle \frac{3y}{4}}\right)}^4$ will be maximum if all the factors are equal
i.e., 2x3=3y4=2x+3y3+4=77=1${\displaystyle \frac{2x}{3}={\displaystyle \frac{3y}{4}={\displaystyle \frac{2x+3y}{3+4}={\displaystyle \frac{7}{7}=1}}}}$
So maximum value of Z = (32)3(43)4(1)3(1)4${\left({\displaystyle \frac{3}{2}}\right)}^3{\left({\displaystyle \frac{4}{3}}\right)}^4(1{)}^3(1{)}^4$ = 278×25681=323${\displaystyle \frac{27}{8}\times {\displaystyle \frac{256}{81}={\displaystyle \frac{32}{3}}}}$

Alternate method:

We partial differentiate the given function w.r.t x and then with y and find the ratio.  Also we partial differentiate 2x+3y = 7 w.r.t x and then with y and find the ratio. Now we equate these two ratio's and find y value interms of x.
⇒$\Rightarrow$3x2y44y3x3=23${\displaystyle \frac{3x^2{y}^4}{4y^3{x}^3}={\displaystyle \frac{2}{3}}}$
⇒$\Rightarrow$3y4x=23⇒yx=89${\displaystyle \frac{3y}{4x}=\frac{2}{3}\Rightarrow {\displaystyle \frac{y}{x}=\frac{8}{9}}}$
⇒y=89x$\Rightarrow y={\displaystyle \frac{8}{9}x}$
Substituting in 2x + 3y = 7 we get x = 32${\displaystyle \frac{3}{2}}$
Now we find value of y as 43${\displaystyle \frac{4}{3}}$
So maximum value of x3.y4$x^3.y^4$ = (32)3.(43)4=323${\left({\displaystyle \frac{3}{2}}\right)}^3.{\left({\displaystyle \frac{4}{3}}\right)}^4={\displaystyle \frac{32}{3}}$

3. If x, y , z are positive reals such that x3y2z4=7$x^3{y}^2{z}^4=7$ then find the minimum value of 2x+5y+3z$2x+5y+3z$
We modify the product to apply AM GM rule.
Consider the product (2x3)3(5y2)2(3z4)4${\left({\displaystyle \frac{2x}{3}}\right)}^3{\left({\displaystyle \frac{5y}{2}}\right)}^2{\left({\displaystyle \frac{3z}{4}}\right)}^4$
Above is the product of nine quantities.
Apply AM ≥$\ge$ GM
⇒(3.2x3+2.5y2+4.3z4)3+2+4≥{(2x3)3.(5y2)2.(3z4)4}$\Rightarrow {\displaystyle \frac{\left(3.{\displaystyle \frac{2x}{3}+2.{\displaystyle \frac{5y}{2}+4.{\displaystyle \frac{3z}{4}}}}\right)}{3+2+4}\ge \left\{{\left({\displaystyle \frac{2x}{3}}\right)}^3.{\left({\displaystyle \frac{5y}{2}}\right)}^2.{\left({\displaystyle \frac{3z}{4}}\right)}^4\right\}}$
⇒(3.2x3+2.5y2+4.3z4)3+2+4≥{(23)3.(52)2.(34)4.x3y2z4}1/9$\Rightarrow {\displaystyle \frac{\left(3.{\displaystyle \frac{2x}{3}+2.{\displaystyle \frac{5y}{2}+4.{\displaystyle \frac{3z}{4}}}}\right)}{3+2+4}\ge {\left\{{\left({\displaystyle \frac{2}{3}}\right)}^3.{\left({\displaystyle \frac{5}{2}}\right)}^2.{\left({\displaystyle \frac{3}{4}}\right)}^4.x^3{y}^2{z}^4\right\}}^{1/9}}$
⇒2x+5y+3z≥9{827.254.81256.7}1/9$\Rightarrow 2x+5y+3z\ge 9{\left\{{\displaystyle \frac{8}{27}.{\displaystyle \frac{25}{4}.{\displaystyle \frac{81}{256}.7}}}\right\}}^{1/9}$
⇒2x+5y+3z≥9{52527}1/9$\Rightarrow 2x+5y+3z\ge 9{\left\{{\displaystyle \frac{525}{2^7}}\right\}}^{1/9}$

4. Find the maximum value of (7−x)4(2+x)5${\left(7-x\right)}^4{\left(2+x\right)}^5$ when x lies between - 2, 7.
To apply any of the above said rules, we first consider that the given terms are positive or not.  7-x, 2+x both are positive between -2, 7
We have to find max. value of (7−x)4(2+x)5${\left(7-x\right)}^4{\left(2+x\right)}^5$ or A4B5$A^4{B}^5$ where A + B = 9.
It will be maximum if (A4)4(B5)5${\left({\displaystyle \frac{A}{4}}\right)}^4{\left({\displaystyle \frac{B}{5}}\right)}^5$ is maximum
Their sum is 4(A4)+5(B5)$4\left({\displaystyle \frac{A}{4}}\right)+5\left({\displaystyle \frac{B}{5}}\right)$ = A + B = 9
For max.product A4=B5=A+B4+5=99=1${\displaystyle \frac{A}{4}={\displaystyle \frac{B}{5}={\displaystyle \frac{A+B}{4+5}={\displaystyle \frac{9}{9}=1}}}}$
So A = 4 and B = 5
Max. product is 4455$4^4{5}^5$

Alternate Method:

We know that

The greatest value of am.bn.cp$a^m.b^n.c^p$, when m, n, p being +ve integers, a+b+c is constant is given by
mm.nn.pp......(a+b+c+...m+n+p+...)m+n+p+..$m^m.n^n.p^p......{\left({\displaystyle \frac{a+b+c+...}{m+n+p+...}}\right)}^{m+n+p+..}$
Therefore max value of the above = 4455(7−x+2+x4+5)4+5=4455(99)4+5=4455