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

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.

• Hint: the second quadratic form is clearly positive definite and associated to a symmetric matrix, whose real eigenvalues describe... – Jack D'Aurizio Jan 27 at 2:49
• You’re looking for a norm of an obvious matrix... – Macavity Jan 27 at 5:52

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

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

• this is the shortest and perfect solution. – Math_centric Jan 29 at 11:22

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.

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.

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).

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

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

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