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3 years ago

9

An entry in the Peach Festival Poster Contest must be rectangular and have an area of 1200 square inches. Furthermore, it's leng

th must be 20 inches longer than it's width. Find the dimensions.

1 answer:

Charra [1.4K]3 years ago

7 0

Answer:

The length is 46.05551275 inches, and the width is 26.05551275 inches.

Step-by-step explanation:

We know that the area must be 1200 square inches. Using this information, we can create an equation, where x is length and y is width:

x*y=1200

We know that its length must be 20 inches longer than its width. Therefore, x=y+20. Using this new information, we can replace 'x' in 'x*y=1200' with 'y+20':

(y+20)*y=1200

$y^{2} +20y=1200$

$y^{2} +20y-1200=0$

I have decided to use the quadratic formula, but you could also factor this equation into the 'intercept' form to determine the roots, which ultimately provides the same answer.

$y=\frac{-b+\sqrt{b^{2} -4ac} }{2a}$

$y=\frac{-(20)+\sqrt{(20)^{2} -4(1)(-1200)} }{2(1)}$

$y=\frac{-(20)+\sqrt{400+4800} }{2}$

$y=\frac{-(20)+\sqrt{5200} }{2}$

$y=\frac{52.11102551 }{2}$

$y=26.05551275$ inches

$x=y+20$

$x=(26.05551275)+20$

$x=46.05551275$ inches

Therefore, the length is 46.05551275 inches, and the width is 26.05551275 inches.

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Answer:

The general solution of the differential equation is $\frac{49y^{2} }{2}-\frac{x^{2} }{2} = c_{3}$

The equation of the solution through the point (x,y)=(7,1) is $y=\frac{x}{7}$

The equation of the solution through the point (x,y)=(0,-3) is $\:y=-\frac{\sqrt{441+x^2}}{7}$

Step-by-step explanation:

This differential equation $\frac{dy}{dx}=\frac{x}{49y}$ is a separable first-order differential equation.

We know this because a first order differential equation (ODE) $y' =f(x,y)$ is called a separable equation if the function $f(x,y)$ can be factored into the product of two functions of <em>x</em> and <em>y</em>

$f(x,y)=p(x)\cdot h(y)$ where<em> p(x) </em>and<em> h(y) </em>are continuous functions. And this ODE is equal to $\frac{dy}{dx}=x\cdot \frac{1}{49y}$

To solve this differential equation we rewrite in this form:

$49y\cdot dy=x \cdot dx$

And next we integrate both sides

$\int\limits {49y} \, dy=\int\limits {x} \, dx$

$\mathrm{Apply\:the\:Power\:Rule}:\quad \int x^adx=\frac{x^{a+1}}{a+1}\\\int\limits {49y} \, dy=\frac{49y^{2} }{2} + c_{1}$

$\int\limits {x} \, dx=\frac{x^{2} }{2} +c_{2}$

So

$\int\limits {49y} \, dy=\int\limits {x} \, dx\\\frac{49y^{2} }{2} + c_{1} =\frac{x^{2} }{2} +c_{2}$

We can subtract constants $c_{3}=c_{2}-c_{1}$

$\frac{49y^{2} }{2} =\frac{x^{2} }{2} +c_{3}$

An explicit solution is any solution that is given in the form $y=y(t)$ . That means that the only place that y actually shows up is once on the left side and only raised to the first power.

An implicit solution is any solution of the form $f(x,y)=g(x,y)$ which means that y and x are mixed (<em>y</em> is not expressed in terms of <em>x</em> only).

The general solution of this differential equation is:

$\frac{49y^{2} }{2}-\frac{x^{2} }{2} = c_{3}$

To find the equation of the solution through the point (x,y)=(7,1)

We find the value of the $c_{3}$ with the help of the point (x,y)=(7,1)

$\frac{49*1^2\:}{2}-\frac{7^2\:}{2}\:=\:c_3\\c_3 = 0$

Plug this into the general solution and then solve to get an explicit solution.

$\frac{49y^2\:}{2}-\frac{x^2\:}{2}\:=\:0$

$\mathrm{Add\:}\frac{x^2}{2}\mathrm{\:to\:both\:sides}\\\frac{49y^2}{2}-\frac{x^2}{2}+\frac{x^2}{2}=0+\frac{x^2}{2}\\Simplify\\\frac{49y^2}{2}=\frac{x^2}{2}\\\mathrm{Multiply\:both\:sides\:by\:}2\\\frac{2\cdot \:49y^2}{2}=\frac{2x^2}{2}\\Simplify\\9y^2=x^2\\\mathrm{Divide\:both\:sides\:by\:}49\\\frac{49y^2}{49}=\frac{x^2}{49}\\Simplify\\y^2=\frac{x^2}{49}\\\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}$

$y=\frac{x}{7},\:y=-\frac{x}{7}$

We need to check the solutions by applying the initial conditions

With the first solution we get:

$y=\frac{x}{7}=\\1=\frac{7}{7}\\1=1\\$

With the second solution we get:

$\:y=-\frac{x}{7}\\1=-\frac{7}{7}\\1\neq -1$

Therefore the equation of the solution through the point (x,y)=(7,1) is $y=\frac{x}{7}$

To find the equation of the solution through the point (x,y)=(0,-3)

We find the value of the $c_{3}$ with the help of the point (x,y)=(0,-3)

$\frac{49*-3^2\:}{2}-\frac{0^2\:}{2}\:=\:c_3\\c_3 = \frac{441}{2}$

Plug this into the general solution and then solve to get an explicit solution.

$\frac{49y^2\:}{2}-\frac{x^2\:}{2}\:=\:\frac{441}{2}$

$y^2=\frac{441+x^2}{49}\\\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}\\y=\frac{\sqrt{441+x^2}}{7},\:y=-\frac{\sqrt{441+x^2}}{7}$

We need to check the solutions by applying the initial conditions

With the first solution we get:

$y=\frac{\sqrt{441+x^2}}{7}\\-3=\frac{\sqrt{441+0^2}}{7}\\-3\neq 3$

With the second solution we get:

$y=-\frac{\sqrt{441+x^2}}{7}\\-3=-\frac{\sqrt{441+0^2}}{7}\\-3=-3$

Therefore the equation of the solution through the point (x,y)=(0,-3) is $\:y=-\frac{\sqrt{441+x^2}}{7}$

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