Figure Out The Mathematics Behind The Famous Golden Arches

The height of the logo is the distance between the origin and the point where the curves touch each other. So you can solve both the equations to get the height. But if you are smart enough to observe that the curves are symmetrically positioned about Y-axis, your life will be a hell lot easier. Coz, this means both the curves touch the Y-axis at same point and that point is their point of intersection. SO simply put x=0 in the equation and get y= -25A. So the height is 25A. Now for the length, observe that each parabola is again symmetric about the vertical line drawn through the point at which they meet X-axis. SO, the length is 2 2 5=20 (If you can't understand this logic, you can also put y=-25A in both equation and compute all the values of x). now, 25A/20=1.05 => A=0.84

For non-zero values of $A$ , it's clear that the length of the arc will be $20$ . With this, we can know the height of the two parabolas, which is $21$ . The height is $21$ when $x=0$ . Then, $21=-A(0-5)^2$ $21=-A.25$ $-A=21/25$ $A=|-0.84|$

So, $A$ is $\boxed{0.84}$ .

at x=0

H=25A

at y=0

L/4=5

H/L=25A/20=1.05

A=0.84

Firstly, we must determine where $y_1 = -A(x-5)^2$ and $y_2 = -A(x+5)^2$ intersect. Using a symmetrical argument (though I will demonstrate that it exists in a moment), there is an intersection point of the two curves somewhere along the axis of $x=0$ . Doing out the algebra when $x=0$ : $y_1 = -A(0-5)^2 \rightarrow y_1 = -25A y_2 = -A(0+5)^2 \rightarrow y_2 = -25A$ Thus the curves intersect at $(0, -25A)$ .

With this in mind, we can find the other x coordinates of the parabolas at that y coordinate. Algebra when $y_1 = y_2 = -25A$ : $-25A = -A(x-5)^2 \rightarrow x = 10 -25A = -A(x+5)^2 \rightarrow x = -10$

The difference between those values of x is the overall length of 20. Since the height (h) to length (l) ratio is 1.05, we find the following: $1.05 = \frac{h}{l} = \frac{h}{20} \rightarrow h = 20 \cdot 1.05 = 21$

From the diagram above, $y = -h$ , and since $y = -25A$ , $A = 21/25$ , or rather, 0.84.

Note that for the parabolas, the $x$ -intercept is $5$ and $-5$ , which means that the length is $20$ . Then, the height is forced to be $21$ , because $\frac {21}{20} = 1.05$ . So, the $y$ -intercept should be $-21$ . But then that implies that $-A\cdot 25 = -21$ , and hence $A = \frac {21}{25} = \boxed {0.84}$ .

clearly we can assume the parabolas in 3rd and 4th quadrant. then y=0 will give the dist of topmost point from the origin (as square of any real no. is always positive and 0is greater than all negative values) it comes out to be x=5 for for the parabola y= -A(x-5)^2 and x=-5 for second one. so the length becomes 2*(5-(-5))=20 for height consider the intersection point of both the parabolas, let it be (x,y). since at intersection it will lie n both the parabolas and hence satisfies both the equations so -A(x-5)^2= -A(x+5)^2 which gives x=0 . therefor value of y at x=0 gives the height of parabola which comes out to be -25A(- sign indicate below the x axis) so 25A/20=1.05 which gives A=0.840

I think the arches would look more right side up if it was expressed in this way:

$y=21-0.84{ (x\pm 5) }^{ 2 }$

but that's just me.

H/L = 1.05 which L = 5 * 4 = 20

then

H = 1.05 * 20 = 21

$y_{1} = -A(x^{2} - 10x + 25) = -Ax^{2} + 10A x - 25A$

$y_{2} = -A(x^{2} + 10x + 25) = -Ax^{2} - 10A x - 25A$

if x = 0 then

$y_{1} = y_{2} = -25A = -21$

$A = 21 / 25 = 0.84$

Vertex of the parabola equations are (5, 0) and (-5, 0). Putting x = 0, we get y = -25A from both equations, it means both cuts y-axis at (0, -25A). We can see from the equations that first one is symmetric respect to the line x = 5, and second one is symmetric respect to x = -5, thus we get that for equation y = -A(x - 5)^2 we find points (0, -5A) and (10, -5A) [because (5, 0) is the midpoint of (0, 0) and (10, 0)]. let height = h, and length = l. So h/l = 1.05 h = 1.05 * l....... Now the base made by a single parabola is = b. So l = 2b. So h = 1.05 * 2 * b = 2.1 * b. Now, y = -h (from the graph of first parabola) and b = x So, -y = 2.1x => y = -2.1 * 10 => y = -21 and y = -25A => -21 = -25A => A = 21/25 => A = 0.84

The equation $f\left( x \right) =-A{ (x-5­) }^{ 2 }$ can be lifted by the height of the logo to make the bottom of the logo correspond with zeroes of the function. $\quad f\left( x \right) =-A{ (x-5­) }^{ 2 }+h=-A{ (x-5) }^{ 2 }+1.05l$

Through expanding and simplifying, this comes to:

$0={ x }^{ 2 }-10x+(25-1.05l{ A }^{ -1 })$

Since the roots of a quadratic are $\frac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a }$ , the difference in x value between quadratic roots is $\frac { -b+\sqrt { { b }^{ 2 }-4ac } }{ 2a } -\frac { -b-\sqrt { { b }^{ 2 }-4ac } }{ 2a } =\quad \frac { \sqrt { { b }^{ 2 }-4ac } }{ a }$ .

The difference in roots should be half the length of the logo, so: $\frac { \sqrt { { b }^{ 2 }-4ac } }{ a } =\quad \frac { \sqrt { 100-4(25-1.05*{ A }^{ -1 }) } }{ 1 } =\sqrt { 4.2l*{ A }^{ -1 } } =\frac { l }{ 2 }$

Simplifying, $\frac { 16.8 }{ l } =A$

We know that l = 20 in the given situation because one fourth of the length of the logo = 5. $\frac { 16.8 }{ 20 } =.84$

I understand this isn't the fastest solution, I like to be more general where I can:)

We squaring $y=-A(x-5)^2=-Ax^2+10Ax-25A$ and displace this graph such that it cross on (0,0) point .

that is $y=-A(x-5)^2+25A$ (1) analogous for $y=-A(x+5)^2+25A$ (2) .

Next we find the solutions of (1) wich are x=0 and x=10 and we can note the distance between 10 and 0 are $\frac{Lenght}{2}$ , analogously for (2).

so, we conclude the Lenght=20 using the ratio (Height/Lenght=1.05) we find the height are 21, this value is the maxima of (1) and (2) then the maximal value of (1) reach when x=5 ( a half of distance between solutions).

Finally: $-A(x-5)^2+25A=-A(5-5)^2+25A=21$

$A= \frac{21}{25}=0.84$

Length=20 (from -10 to +10). Height= 25A. Given, 25A/20=1.05. This gives A=0.84

Find the y-intercept to determine the height of the logo, x=0, y=-25A. Thus, the height is 25A. Then substitute this value to the two equation will get x=-10 for left and x=10 for right, Thus, the length=20. From question, 25A/20=1.05, solving the equation then we get A=0.84

The two curves intersect at $x = 0$

at $x = 0$ :

$h = y$

$h=-25A$ -----> $(1)$

$w = 4(x-5)$

$w = -20$ -----> $(2)$

$\frac{h}{w} = \frac{-25A}{-20} = 1.05$

solving this equation makes $A = 0.84$

In the graph, we notice that when x = 0 => y = -H => -H = -25A => H = 25A Since H/L = 1.05 => L = 500/21 A Moreover, when x = -L/2 => y = -H Therefore 25 = (250/21A - 5)^2 => A = 0.84

height/length=1.05; for x=5 or -5, y=0. so for 5 units either way (+x or -x), you get to midway of either arch (y=0). total length of both archs is 5 added 4 times (4 half arches, one half arch is 5 ) length=20 , so height is 1.05*20=21; for x=0, y=25A=21=height. so A is 0.84

Since the parabolic curves are symmetric around their vertices which are (-5,0) and (5,0) and intersect when x=0, the length of the base is always 20. The height is the y-value of their intersection. If A were 1, the height would be 25, and since the length to height ratio is 1.05, we want the height to be 1.05 * (length=20)= 21. Therefore, A = 21/25 = .84.

Using the same scale as the 5s in the formula, the "peaks" occur at x=+/-5 and the width is 20. the "height" is expressed as a negative so that the curves are the correct shape so the initial - can be ignored for the purposes of this solution.

Hence: A(5^2) = 1.05 x 20 A = 21/25 = 0.84