Let's do some calculus! (46)

Calculus Level 5

If the maximum value of

f ( x , y , z ) = 2 x 2 + 3 y − 4 z \large f(x,y,z) = 2x^2 + 3y - 4z

subject to the constraints

{ 2 x − 4 y − 2 z = 5 x 2 + y 2 = 1 \begin{cases} 2x-4y-2z=5 \\ x^2+y^2 = 1 \end{cases}

is equal to M M , then evaluate ⌊ M ⌋ \left\lfloor M \right\rfloor .

Note: It is advised to take the aid of a calculator for mathematical computations in this problem.

For more problems on calculus, click here .


The answer is 22.

Tapas Mazumdar by Tapas Mazumdar , 20, India

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1 solution

Chew-Seong Cheong
May 27, 2017

Since x 2 + y 2 = 1 x^2+y^2=1 , we can substitute x = cos ⁡ θ x=\cos \theta and y = sin ⁡ θ y = \sin \theta , then from 2 x − 4 y − 2 z − 5 2x-4y-2z-5 , we have 2 z = 2 cos ⁡ θ − 4 sin ⁡ θ − 5 2z = 2\cos \theta - 4 \sin \theta - 5 and:

f ( x , y , z ) = 2 x 2 + 3 y − 4 z = 2 cos ⁡ 2 θ + 3 sin ⁡ θ − 2 ( 2 cos ⁡ θ − 4 sin ⁡ θ − 5 ) = 2 cos ⁡ 2 θ + 11 sin ⁡ θ − 4 cos ⁡ θ + 10 \begin{aligned} f(x,y,z) & = 2x^2+3y-4z \\ & = 2\cos^2 \theta + 3\sin \theta - 2(2\cos \theta - 4 \sin \theta - 5) \\ & = 2\cos^2 \theta + 11 \sin \theta - 4\cos \theta + 10 \end{aligned}

By plotting we find that M ≈ 22.031 M \approx 22.031 ⟹ ⌊ M ⌋ = 22 \implies \lfloor M \rfloor = \boxed{22} .

4 Helpful 0 Interesting 0 Brilliant 0 Confused

Let's say we want to proceed from here

f ( θ ) = 2 cos ⁡ 2 θ + 11 sin ⁡ θ − 4 cos ⁡ θ + 10 f(\theta) = 2 \cos^2 \theta + 11 \sin \theta - 4 \cos \theta + 10

without plotting graph. How can we do that?

Tapas Mazumdar - 4 years ago

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I was thinking of solving for d f d θ = 0 \dfrac {df}{d\theta} = 0 . Use Newton's method (numerical) to find θ \theta , then find f ( θ ) f(\theta) . It would be very long and it is still numerical.

Chew-Seong Cheong - 4 years ago

I also proceed up to this but I got s degree 4 polynomial in x which is no solvable by normal method

So, do u knows any other ways ?

Kushal Bose - 4 years ago

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Well this method of taking the polar form is very efficient, way better than what I has used (Lagrange multipliers) and by that method too I ended up in a biquadratic which I had to take aid of computer to solve. So, yes a graph plot is a good way.

Tapas Mazumdar - 4 years ago

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@Tapas Mazumdar – I doubt you can get a biquadratic expression. If what you said is true, then it's probably the most elegant solution here.

Can you share with us on how you get this result?

Pi Han Goh - 4 years ago

I used Lagrange Multipliers to get a system of nonlinear equations. Then used Multivariate Newton Raphson.

Steven Chase - 4 years ago

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@Steven Chase – Same, Langrange was giving me a quartic equation for the λ 2 \lambda_2 term, that was better to solve.

Tapas Mazumdar - 4 years ago

@Steven Chase – I think you meant Hessian Matrix. It's impossible to prove the maximality via Lagrange Multipliers alone.

Pi Han Goh - 4 years ago

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@Pi Han Goh – I took it for granted that it would be a maximum, since the question said so :). The Lagrange Multipliers gives a nonlinear system. To solve the nonlinear system, you can use the matrix (multivariate) version of Newton Raphson. But yes, strictly speaking, even when you have a solution, you don't know whether it's a max, min, or inflection point.

Steven Chase - 4 years ago

The maximum value occurs when θ = 2 π k + 2 arctan ⁡ [ 4 11 + 1 2 64 121 + 2 − 144 + 64659 3 3 3 2 / 3 − 2 × 1 1 2 / 3 3 ( 64659 − 144 ) 3 + 1 2 128 121 − 2 − 144 + 64659 3 3 3 2 / 3 + 2 × 1 1 2 / 3 3 ( 64659 − 144 ) 3 + 1024 1331 64 121 + 2 − 144 + 64659 3 3 3 2 / 3 − 2 × 1 1 2 / 3 3 ( 64659 − 144 ) 3 ] \theta =2\pi k + 2 \arctan \left [\dfrac4{11} + \dfrac12 \sqrt{ \dfrac{64}{121} + \dfrac{2\sqrt[3]{-144 + \sqrt{64659}}}{33^{2/3}} - \dfrac{2\times11^{2/3}}{\sqrt[3]{3(\sqrt{64659}-144)}}} + \dfrac12 \sqrt{\dfrac{128}{121} - \dfrac{2\sqrt[3]{-144 + \sqrt{64659}}}{33^{2/3}} + \dfrac{2\times11^{2/3}}{\sqrt[3]{3(\sqrt{64659}-144)}} + \dfrac{1024}{1331\sqrt{ \dfrac{64}{121} + \dfrac{2\sqrt[3]{-144 + \sqrt{64659}}}{33^{2/3}} - \dfrac{2\times11^{2/3}}{\sqrt[3]{3(\sqrt{64659}-144)}}} } } \right ] for any integer k k .

This value can be determined using half angle tangent substitution followed by the quintic formula .

Pi Han Goh - 4 years ago

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Woah! That's a really big expression! This is why I recommended using computational aid in this step.

Tapas Mazumdar - 4 years ago

Wow, you are real good.

Chew-Seong Cheong - 4 years ago

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@Chew-Seong Cheong – Thanks for the compliment.... although this is just pure bashing without any real creativity.

Here's an alternative approach (without half angle substitution)

f ( θ ) f(\theta) can also be written as f ( θ ) = 12 − 2 sin ⁡ 2 x + 11 sin ⁡ x − 4 cos ⁡ x f(\theta) = 12 - 2\sin^2 x + 11\sin x - 4\cos x .

Then f ′ ( θ ) = 11 cos ⁡ x + 4 sin ⁡ x − 4 sin ⁡ x cos ⁡ x f'(\theta) = 11\cos x + 4\sin x - 4\sin x \cos x . When at critical point, f ′ ( θ ) = 0 f'(\theta) = 0 , so

4 sin ⁡ x cos ⁡ x − 4 sin ⁡ x = 11 cos ⁡ x 4 sin ⁡ x ( 1 − cos ⁡ x ) = 11 cos ⁡ x 16 sin ⁡ 2 x ( 1 − cos ⁡ x ) 2 = 121 cos ⁡ 2 x 16 ( 1 − cos ⁡ 2 x ) ( 1 − cos ⁡ x ) 2 = 121 cos ⁡ 2 x 4\sin x \cos x - 4\sin x = 11\cos x \\ 4\sin x (1 - \cos x) = 11\cos x \\ 16 \sin^2 x (1 - \cos x)^2 = 121 \cos^2 x \\ 16 (1-\cos^2 x)(1- \cos x) ^2= 121 \cos^2 x

Then by the quintic formula, we can solve for cos ⁡ x \cos x , which is an algebraic number. What's left is to prove that f ′ ′ ( θ ) < 0 f''(\theta) < 0 and we're done.

To Tapaz: A good (albeit, super duper extremely tedious) follow up question to ask is "Find the exact closed form of the maximum value of f ( θ ) f(\theta) ".

Pi Han Goh - 4 years ago

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