Why isn't it 0

Let $a=5x+10y$ . Then the expression becomes $(3a+20)^2+(4a+15)^2$ $=25a^2+240a+625$ $=(5a+24)^2+49\ge 49.$ Thus, the minimum is $\boxed{49}$ .

The minimum is 49: To find the minimum, we can take the partial derivatives of $f(x) = (15x + 30y + 20)^2 + ( 20x + 40y + 15)^2$ with respect to x and y and set them equal to zero. This gives: $\frac{\partial f}{\partial x} = 2 * 15 * (15x + 30y + 20) + 2 * 10 * ( 20x + 40y + 15) = 0$ $\frac{\partial f}{\partial y} = 2 * 30 * (15x + 30y + 20) + 2 * 40 * (20x + 40y + 15) = 0$ Further simplification yields a single equation: $x + 2y = -\frac{24}{25}$ or $2y = -x - \frac{24}{25}$ Plugging this into the original function gives $f(x) = \left( \frac{28}{5}\right)^2 + \left( -\frac{21}{5} \right)^2 = \frac{1225}{25} = 49$

We can partially factorize $15x+30y+20=15(x+2y)+20$ and $20x+40y+15=20(x+2y)+15$ . Since $x,y$ ranges over all real values, $x+2y$ can take on all real values; we can therefore replace $x+2y$ as $z$ . We need to find the minimum of $(15z+20)^2+(20z+15)^2$ . Expanding and completing the square, we get $(25z+24)^2+49$ . The minimum $(25z+24)^2$ can be is $0$ ; therefore, the minimum the whole thing can be is $\boxed{49}$ .

• Let's first take 25 out from both terms

$[5(3x+6y+4)]^2+[5(4x+8y+3)]^2$

$25(3x+6y+4)^2+25(4x+8y+3)^2$

$25[(3x+6y+4)^2+(4x+8y+3)^2]$

• If we use to do factorizing, we can easily see the factor " $x+2y$ " inside the terms. So let's factorize it.

$25{[3(x+2y)+4]^2+[4(x+2y)+3]^2}$

• But it seems too much brackets there, so let's call $x+2y=k$ to make it simpler.

$25[(3k+4)^2+(4k+3)^2]$

• It is now clear that minimum value we were looking for is $25 \times \min((3k + 4)^{2} + (4k + 3)^{2})$ . Using the formula of extreme point of quadratic function, gives us $\min((3k+4)^2+(4k+3)^2) = \frac{49}{25}$ . And so, the final answer is $25 \times \frac{49}{25} = \boxed{49}$

Choose x + 2y = z So the given equation becomes: (15z + 20)^2 + (20z + 15)^2 = F(z) (Say) i.e. F(z) = 25z^2 + 48z + 9 .............(i) Differentiating F w.r.t. z & computing F'(z) = 0 we get z = - 24/25 Also F''(- 24/25) > 0 which means that the function attains it minima at z = - 24/25 So putting z = - 24/25 in (i) we get 49

The reason I could solve this problem was because the x and y coefficients have a common ratio (that is x:2y).

We find the lowest common multiple of the two expressions and we find that it is $60x+120y$

Let $z = 60x+120y$ then we can rewrite the expression as $\left(\frac{z}{4} + 20\right)^{2} + \left(\frac{z}{3} + 15\right)^{2}$ .

Now we expand the expression and differentiate:

$0 = \frac{d}{dz}\left(\frac{25}{144}z^{2} + 20z + 625\right) = \frac{50}{144}z + 20$

$\frac{50}{144}z = -20 \implies z = -\frac{288}{5}$

Therefore we substitute this value of z into our formula (as it will give us the minimum value):

$\left(-\frac{288}{20}+20\right)^{2} + \left(-\frac{288}{15}+15\right)^{2} = 49$

So our answer is $\fbox{49}$

Observe that

$(15x+30y+20)^{2}+(20x+40y+15)^{2}=(25x+50y+24)^{2}+49$ .

But

$(25x+50y+24)^{2}\ge0$ .

Hence,

$(25x+50y+24)^{2}+49\ge49$ .

Which imlies that the minimum must be $\boxed{49}$

$(15x + 30y + 20)^2 + (20x + 40y + 15)^2$

Factorizing:

$(5(3x + 6y + 4))^2 + (5(4x + 8y + 3))^2$

$= 25(3x + 6y + 4)^2 + 25(4x + 8y + 3)^2$

$= 25( (3x + 6y + 4)^2 + (4x + 8y + 3)^2 )$

In both equations, the $x$ and (y) can be factorized into $(x + 2y)$ , giving

$25( (3(x + 2y) + 4)^2 + (4(x + 2y) + 3)^2 )$

We don't need to know the individual values of $x$ and $y$ , only the value of $x + 2y$ . Changing $x + 2y$ to $a$ :

$25( (3a + 4)^2 + (4a + 3)^2 )$

Squaring:

$25( 9a^2 + 24a + 16 + 16a^2 + 24a + 9 )$

$= 25( (25a^2 + 48a + 25)$

We now have a straightforward quadratic equation. To find the vertex, we find two $a-values$ that give the same result. $25a^2 + 48a = 0$

$a(25a + 48) = 0$

So $a = 0$ or $25a + 48 = 0$ this being $a = -\frac{48}{25}$ . The vertex is at the value of $a$ halfway between these two so $a = - \frac {\frac{48}{25} }{2} = - \frac{24}{25}$

Putting this into the equation, we have:

$25(25(- \frac{24}{25})^2 + 48( - \frac{24}{25}) + 25)$

$25(25(\frac{24}{25})(\frac{24}{25}) - 48(\frac{24}{25} )+ 25)$

$25( \frac{24^2}{25} - \frac{48(24)}{25} + 25)$

$= (24^2 - 48(24) + 25^2)$

$= (25^2 - 24^2)$

$= (25 + 24)(25 - 24)$

= $49$

$$\left(15x+30y+20\right)^2+\left(20x+40y+15\right)^2$$ $$=\left[15(x+2y)+20\right]^2+\left[20(x+2y)+15\right]^2$$ $$= \left(10^2+20^2\right)(x+2y)^2 + 2.600(x+2y)+625$$ $$= \left[25(x+2y)\right]^2 + 2.25.24(x+2y)+576+49$$ $$= \left[25(x+2y)+24\right]^2 + 49 \ge 49$$ $$\boxed{\text{My solution is 49}}$$

Note that $(15x+30y+20)^2+(20x+40y+15)^2 = \\ 625 x^2+2500 x y+1200 x+2500 y^2+2400 y+625 = \\ (25x + 50y + 24)^2 + 49$

Hence $\boxed{49}$ is the answer.

We can see that $(15x+30y+20)^2+(20x+40y+15)^2=(25x+50y+24)^2+49$ Since, $x,y$ ranges over all real numbers there are such pairs of $(x,y)$ for which $(25x+50y+24)^2=0$ .

So, the minimum value of $(15x+30y+20)^2+(20x+40y+15)^2=(25x+50y+24)^2+49=0+49=49$

Consider $a = 17.5x + 35y + 17.5$ and $b = 2.5x +5y -2.5$

Let F be the function to minimise We see that $F = (a+b)^{2} + (a-b)^{2}$

Also, by inspection $a = 7b + 35$

Substituting for $a$ and simplifying, we get $F = 100 ( b^{2} + \frac{98}{10}b + \frac{49}{2} )$

$=$ $100( (b+ \frac{49}{10})^{2} + \frac{49}{2} - \frac{2401}{10} )$

This is minimum at $b = - 49/10$ (to make the square 0). Since $b$ is linear in $x$ and $y$ , it can take this value.

Hence $F_{min} = 100( \frac{49}{2} - \frac{2401}{100} )$ which is 49

Let n = 5 x +10 y

f( n ) = (3 n +20)^2+(4 n +15)^2 = 25 n ^2+240 n +625

To find the minimum of a quadratic, we must find where the slope is 0, which is determined by solving for the derivative f'( n ) = 50 n +240; f'( n ) = 0 when n = -4.8

f(-4.8) = 25(-4.8)^2+240(-4.8)+625 = 576-1152+625 = 49

Okay so we start by expanding everything...

$(15x+30y+20)^2 + (20x+40y+15)^2 \implies$

$625x^2 + 2500xy + 1200x + 2500y^2 + 2400y + 625$

Then we can subtract 49 (and then add it back) to make it factorable.

$\implies 625x^2 + 2500xy + 1200x + 2500y^2 + 2400y + 576 + 49$ $\implies (25x + 24 + 50y)^2 + 49$

Well making the square value zero, we get $\boxed{49}$ as our final answer.

In order to find the minimum value of this function we can find where its derivative is equal to zero.

Expanding (15x+30y+20)^2 +(20x+40y+15)^2 out we get

625x^2 + 1200x + 2500xy + 2400y + 2500y^2 + 625

The work is kind of messy so I'm going to leave that out.

We want to find the derivative with respect to x and y. They turn out to be the same equation to work with when d/dx and d/dy = 0

So...

d/dx = 2(625x) + 1200 + 2500y = 1250x + 2500y + 1200 = 50(25x + 50y + 24)

Setting this equal to 0 we get

25x + 50y + 24 = 0

We need to find the intercept values of x and y when 25x + 50y = -24 y = (-24 - 25x)/(50)

x-intercept = -24/25 y-intercept = -24/50

Plugging x into the original equation and 0 for y we get 49. We also get 49 for substituting the y-intercept value and 0 for x.

(15(-24/25) + 20)^2 + (20(-24/25) + 15)^2 = 49

(30(-24/50) + 20)^2 + (40(-24/50) + 15)^2 = 49

É possivel resolver essa questão usando conhecimento de "matemática avançada". Basta chamar a expressão dada de z(x,y) e fazer as derivadas parciais em relação a x e y depois igualá-las à zero, após isso um sistema linear será encontrado e encontrar os valores de x e y que darão o valor mínimo de z(x,y) que serão x= -0.96 e y= 0.

Let the variable a = 15x + 30y. Then 20x + 40y is 4/3 a, and I am minimizing the sum of two quadratics in a, which is itself a quadratic in a. Since the coefficient of the quadratic term is positive, I know I'll find its minimum on the quadratic's axis of symmetry. The axis of symmetry is at a = -72/5. Putting this value of a into the quadratic gave me a minimum of 49.

TAKE 15 AND 20 OUT AND USE DIFFERENTIATION TO SOLVE PROBLEM

If you let x + 2y = z, then the expression becomes (15z+20)^2 + (20z+15)^2 which is 625z^2+1200z+625 Differentiate to get 1250z +1200 and the minimum value of this occurs when z = -24/25 Substitute back into expression to get value of 49

$(15x+30y+20)^2+(20x+40y+15)^2$

$=5^2\{(3x+6y+4)^2+(4x+8y+3)^2\}$

$=5^2\{3^2(x+2y)^2+4^2+24*(x+2y)+4^2(x+2y)+3^2+24*(x+2y)\}$

$=5^2\{5^2(x+2y)^2+48(x+2y)+5^2\}$

$=5^4\{(x+2y+{24 \over 5^2})^2+1-{24^2 \over 5^4}\}$

The minimum value is when $x+2y+{24 \over 5^2}$ is zero. So the minimum value is $5^4(1-{24^2 \over 5^4}) = 25^2-24^2 = 49$ .

(15x+30y+20)^2 + (20x+40y+15)^2

= (15(x+2y)+20)^2 + (20(x+2y(+15)^2

(x+2y = a)

=(15a+20)^2 + (20a+15)^2

=625a^2 + 1200a + 625

= 25(25a^2 + 48a + 25)

f(x) = 25x^2 + 48x +25

f '(x) = 50x+48 = 0

x = -48/50

f ''(x) = 50 (is positive, hence the function only has only minima points - hence x=-48/50 is the minimum point) (that's how we say it in my language so it may sound strange)

if a=-48/50 then we got 25(23.04-46.08+25) = 49