Maximizing $ (4a-3b)^2+(5b-4c)^2+(3c-5a)^2$, such that $a^2+b^2+c^2=1 $

Question

If $$a^2+b^2+c^2=1 $$here a,b,c are the real numbers then find the maximum value of $$ (4a-3b)^2+(5b-4c)^2+(3c-5a)^2$$

I tried to think with vectors, that is direction cosines of lines. But then the expression is not getting simplified.

Answer 1

Use the technique of Lagrange multipliers to solve.

Essentially, if $$f(a,b,c) = a^2 + b^2 + c^2,$$ $$g(a,b,c) = (4a-3b)^2+(5b-4c)^2+(3c-5a)^2,$$ then find the values $(a,b,c)$ such that $\vec \nabla f$ is parallel to $\vec \nabla g$.

Then, by a bit of testing on your solutions, select the one that satisfies $f(a,b,c) = 1$ and is a maximum of $g(a,b,c)$ (as opposed to a minimum or saddle point, for instance).

Answer 2

try opening the brackets. square terms with coeeficents combine them replace $$ a^2+b^2$$ type terms with $$c^2$$.

you have $$ 50-(3a+4b+5c)^2$$

Answer 3

Using the method of Lagrange multipliers, consider $$F= (4a-3b)^2+(5b-4c)^2+(3c-5a)^2+\lambda(a^2+b^2+c^2-1) \tag 1$$ Computing the partial derivatives $$\frac{\partial F}{\partial a}=8 (4 a-3 b)-10 (3 c-5 a)+2 a \lambda \tag 2$$ $$\frac{\partial F}{\partial b}=-6 (4 a-3 b)+10 (5 b-4 c)+2 b \lambda\tag 3$$ $$\frac{\partial F}{\partial c}=6 (3 c-5 a)-8 (5 b-4 c)+2 c \lambda\tag 4$$ $$\frac{\partial F}{\partial \lambda}=a^2+b^2+c^2-1\tag 5$$

From $(2)$, $b=\frac{1}{12} (a \lambda +41 a-15 c)$; plug in $(3)$ to get $c=\frac{1}{15} (a \lambda +25 a)$ which makes $b=\frac{4 }{3}a$; plug $b$ and $c$ in $(4)$ to get $$\frac{2}{15} a \lambda (\lambda +50)=0 \tag 6$$ So, we have three cases : $a=0$ , $\lambda=0$, $\lambda=-50$.

The first case $a=0$ can be discarded since it would make $a=b=c=0$ which does not satisfy the constraint.

The second case $\lambda=0$ would make $c=\frac{5 }{3}a$ which, in turn, would make $\frac{50 }{9}a^2=1$ from the constraint and then the value of the expression to maximize would just be $0$.

So, what is left is the case $\lambda=-50$ which makes $c=-\frac{5 }{3}a$ which gives again $\frac{50 }{9}a^2=1$ from the constraint. The expression to maximize if then $\frac{a^2 \lambda ^2}{9}$ with $\lambda=-50$ and $\frac{50 }{9}a^2=1$ which then gives a maximum vzlue of $50$.

I let you finishing what could be required.

Answer 4

We need to find a minimal value of $k$ for which the following inequality is true for all reals $a$, $b$ and $c$. $$(4a-3b)^2+(5b-4c)^2+(3c-5a)^2\leq k(a^2+b^2+c^2)$$ or $$(k-41)a^2+(k-34)b^2+(k-25)c^2+24ab+40bc+30ac\geq0$$ or $$(k-41)a^2+6(4b+5c)a+(k-34)b^2+40bc+(k-25)c^2\geq0,$$ for which we need $$k>41$$ and $$9(4b+5c)^2-(k-41)((k-34)b^2+40bc+(k-25)c^2)\leq0$$ or $$(k^2-75k+1250)b^2+(k^2-66k+800)c^2+40(k-50)bc\geq0$$ or $$(k-50)((k-25)b^2+(k-16)c^2+40bc)\geq0.$$ But $$(k-25)b^2+(k-16)c^2+40bc\geq16b^2+25c^2+40bc=(4b+5c)^2\geq0,$$ which gives $$k\geq50.$$

For $k=50$ we obtain $$9a^2+6(4b+5c)a+(4b+5c)^2\geq0$$ or $$(3a+4b+5c)^2\geq0,$$ which gives that the equality occurs for $$3a+4b+5c=0$$ and $$a^2+b^2+c^2=1,$$ which says that $50$ is a maximal value.

Answer 5

Assume $\vec{A}=a\hat{i}+b\hat{j}+c\hat{k}$ and $\vec{B}=3\hat{i}+4\hat{j}+5\hat{k}$

$$\bigg|\vec{A}\times \vec{B}\bigg|^2=\bigg|\vec{A}\bigg|^2\bigg|\vec{B}\bigg|^2-\bigg(\vec{A}\cdot\vec{B}\bigg)^2$$

$$(4a-3b)^2+(5b-4c)^2+(3c-5a)^2=50-(3a+4b+5c)^2\leq 50$$