maximum value of $ac+bd$

Question

If $a,b,c,d\in \mathbb{R}$ and $a^2+b^2\leq 2$ and $c^2+d^2\leq 4.$ Then maximum value of $ac+bd$ is

what i try

$(ac+bd)^2+(ad-bc)^2=(a^2+b^2)(c^2+d^2)$

$\underbrace{(ac+bd)^2}_{\max}=\underbrace{(a^2+b^2)(c^2+d^2)}_{\max}-\underbrace{(ad-bc)^2}_{\min}$

$(ac+bd)^2\leq 8\Rightarrow (ac+bd)\leq 2\sqrt{2}$

but answer is $3$

How do i get right answer help me please

Answer 1

You are correct that the maximum value is $2\sqrt 2<3$. As another answer (now deleted; it applied AM-GM separately to $a^2+c^2$ and $b^2+d^2$) points out, you can easily get an upper bound of $3$ using AM-GM, but this bound can't actually be attained, so isn't the correct maximum.

For equality in the AM-GM bound, you would need $a=c$ and $b=d$, but this is clearly impossible since $a^2+b^2\neq c^2+d^2$.

Answer 2

$2 \sqrt2 $ is the correct maximum value.The value $3$ which is larger than $2 \sqrt2 $ is never attained. The bound $2 \sqrt 2$ can also be obtained by Cauchy - Schwarz inequality.

Answer 3

One way of obtaining the maximum value is by doing maximisation with constraints. First observe that $\nabla_{a,b,c,d} (ac+bd)\neq \overrightarrow{0} \text{ for all } a,b,c,d\neq 0$. As it's a continuous function and the domain we're interested in is compact (it's the intersection of two perpendicular cylinders), the maximal value has to be reached somewhere in the domain. But for it to be reached in the interior of the domain, the gradient would have to become 0 at that position, which we observed not to happen. So we have to check the boundary by applying Lagrange multipliers. On the boundary both constraints $a^2+b^2=2$ and $c^2+d^2=4$ are fulfilled. So we add them with Lagrange multipliers and calculate the gradient: $\nabla_{a,b,c,d,\lambda_1,\lambda_2}(ab+cd+\lambda_1(a^2+b^2-2)+\lambda_2(c^2+d^2-4))=\begin{pmatrix}c-\lambda_1\cdot 2a\\d-\lambda_1\cdot2b\\ a - \lambda_2\cdot 2c\\ b - \lambda_2\cdot 2d\\a^2+b^2-2\\c^2+d^2-4\end{pmatrix}$. So now we want this to be zero. The first two lines give $\lambda_1=\frac{c}{2a}=\frac{d}{2b}$, the second two lines give $\lambda_2=\frac{a}{2c}=\frac{b}{2d}$, the third two lines represent our constraints. We're not interested in the exact values of the multipliers, so we just take the second equalities, which are equivalent for both. Solving this equality for d yields $d=\frac{bc}{a}(*)$. This we can insert into the constraint and obtain $c^2+d^2=c^2(1+\frac{b^2}{a^2})=4$. This means $c^2(a^2+b^2)=c^2 \cdot 2=4a^2$ and $c^2=2a^2$. With our equality from above this also means $d^2=2b^2$. We cannot obtain more, but this is already enough, as $(ac+bd)^2=(ac)^2+2acbd+(bd)^2=$ $=(ac)^2+(ad)^2+(bc)^2+(bd)^2=a^2(c^2+d^2)+b^2(c^2+d^2)=4(a^2+b^2)=8$ using the condition $ad=bc$, equivalent to $(*)$ from above. So the maximum value is the positive choice leading to $(ac+bd)=\sqrt{8}=2\sqrt{2}$.

Answer 4

Let $a = r_1 \cos p, b = r_1 \sin p, c = r_2 \cos q$, and $ d = r_2 \sin q$, where $0 < r_1 ≤ \sqrt{2}$ and $0 < r_2 ≤ 2$. This satisfies the inequalities mentioned in the problem.

Then:

$$ac+bd = r_1 r_2\cos p \cos q + r_1 r_2 \sin p \sin q = r_1 r_2 ( \cos p \cos q + \sin p \sin q) = r_1 r_2 \cos(p-q)$$

Since $\cos x ≤ 1$ for all real $x$, the maximum value is $\sqrt{2} \cdot 2 \cdot 1 = 2 \sqrt{2}$.

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Sidenote: This is only one choice of variables. Another choice of variables such as $a = r_1 \sin p, b = r_1 \cos p, c = r_2 \cos q, d = r_2 \sin q$ leads to the same result. Try it yourself!

Answer 5

If you are already comfortable with complex numbers and their basic properties you can also proceed as follows:

• Let $z = a+ib$, $w=c+id$
• $\Rightarrow |z| \leq \sqrt{2}$ and $|w| \leq 2$

It follows $ ac+bd =\Re(z\bar w)\leq |z\bar w| = |z||w| \leq 2\sqrt{2} $

The maximum is reached, for example, for $z=\sqrt 2$ and $w=2$.