The length of a rectangle is 6 centimeters less than twice its width. The perimeter of the rectangle is 48 cm. What are the

Mathematics

1 answer:

fomenos3 years ago

6 0

Answer:

the length is 12 and the width is 9.

Step-by-step explanation:

The first step is always to assign letters to the variables. We can call length L and width W.

Since all units are in cm, unit conversion is unnecessary.

Next, set up some equations. L=2W-6 because of the first sentence of the problem. Length times width equals area, so L*W=108.

Now that there are two equations and two variables, the next step is to solve the system. I'm going to solve it by plugging in the right side of the first equation in for L in the second equation. This gives me (2W-6)*W=108. Distributing the W gives me 2W^2-6W=108, or W^2-3W-54=0. Using the quadratic formula, I get that W=9. Using the first equation, L=2*9-6=12. So the length is 12 and the width is 9.

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Given:

The diameter of the right cylinder is 2x cm.

The total surface area is 96 cm cube.

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$\begin{gathered} r=\frac{d}{2} \\ r=\frac{2x}{2} \\ r=x\text{ cm} \end{gathered}$

The total surface area is,

$\begin{gathered} S=2\pi rh+2\pi(r)^2 \\ 96=2\pi xh+2\pi(x^2) \\ h=\frac{96-2\pi(x^2)}{2\pi x} \end{gathered}$

Volume is,

$\begin{gathered} V=\pi(r)^2h \\ =\pi(x^2)\frac{96-2\pi(x^2)}{2\pi x} \\ =\frac{x(96-2\pi(x^2)}{2} \end{gathered}$

Now, differentiate with respect to x,

$\begin{gathered} \frac{dV}{dx}^{}=\frac{d}{dx}(\frac{x(96-2\pi(x^2)}{2}) \\ =\frac{d}{dx}\mleft(x\mleft(-\pi x^2+48\mright)\mright) \\ =\frac{d}{dx}\mleft(x\mright)\mleft(-\pi x^2+48\mright)+\frac{d}{dx}\mleft(-\pi x^2+48\mright)x \\ =1\cdot\mleft(-\pi x^2+48\mright)+\mleft(-2\pi x\mright)x \\ =84-3\pi(x^2)\ldots\ldots\ldots\ldots\text{.}(1) \end{gathered}$

Now,

$\begin{gathered} \frac{dV}{dx}=0 \\ 84-3\pi(x^2)=0 \\ x^2=\frac{16}{\pi} \\ x=\sqrt[]{\frac{16}{\pi}} \end{gathered}$

Now, differentiate (1) with respect to x again,

$\begin{gathered} \frac{d^2V}{dx^2}=\frac{d}{dx}(84-3\pi(x^2)) \\ =-6\pi x \\ At\text{ x=}\sqrt[]{\frac{16}{\pi}} \\ \frac{d^2V}{dx^2}=-6\pi\sqrt[]{\frac{16}{\pi}}$

Since, the double derivative is negative.