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qaws [65]
3 years ago
8

Katherine wants to construct a small box with a volume of 20 cubic inches with the following specifications. The length of a box

is five more than its width. Its depth is one less than its width. What are the dimensions of the box in simplest radical form and rounded to the nearest hundredth? Only and algebraic solution will receive full credit.
Mathematics
2 answers:
pochemuha3 years ago
5 0
V=LWH=20

L is 5 more than W
V=5+W

H is 1 less than W
H=-1+W

easy, sub 5+W for L and -1+W for H in LWH=V=20

20=(5+W)(W)(-1+W)
expand
20=W^3+4W^2-5W
minus 20 both sides
W^3+4W^2-5W-20=0
either factor or graph to find the possible values for W
(W+4)(W^2-5)=0
set each to zero

W+4=0
W=-4
impossible, measures cannot be negative
W^2-5=0
add 5
W^2=5
sqrt both sides
W=√5

sub back

L=5+W
L=5+√5
aprox
7.24 rounded

H=-1+W
H=-1+√5
aprox
1.24 rounded
the dimentions are
√5 by (5+√5) by (-1+√5) or aprox
2.24in by 7.24in by 1.24in
Arada [10]3 years ago
3 0

Answer:

Width =√5,length =√5+5 ,Depth =√5-1.

Step-by-step explanation:

volume of box = l.w.h ,where l is length ,w is width and h is the depth  of the box.

It is given  l= w+5

                h= w-1.

Substituting l and h values we have

V= (w+5).w.(w-1)

Or 20 = w(w+5)(w-1) [ volume is 20 cubic in)

20 =w(w^{2} +4w-5)

20=w^{3} +4w^{2} -5w

or,w^{3} +4w^{2} -5w-20=0

Taking common factor of both pairs we have:

w^{2} (w+4)-5(w+4)=0

(w^{2} -5)((w+4)=0

w^{2} -5=0  or w+4=0

w=±√5 or w=-w

Length can not be negative so

w=√5

l=w+5=√5+5

h=w-1=√5-1

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2 years ago
If a and b are positive numbers, find the maximum value of f(x) = x^a(2 − x)^b on the interval 0 ≤ x ≤ 2.
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Answer:

The maximum value of f(x) occurs at:

\displaystyle x = \frac{2a}{a+b}

And is given by:

\displaystyle f_{\text{max}}(x) = \left(\frac{2a}{a+b}\right)^a\left(\frac{2b}{a+b}\right)^b

Step-by-step explanation:

Answer:

Step-by-step explanation:

We are given the function:

\displaystyle f(x) = x^a (2-x)^b \text{ where } a, b >0

And we want to find the maximum value of f(x) on the interval [0, 2].

First, let's evaluate the endpoints of the interval:

\displaystyle f(0) = (0)^a(2-(0))^b = 0

And:

\displaystyle f(2) = (2)^a(2-(2))^b = 0

Recall that extrema occurs at a function's critical points. The critical points of a function at the points where its derivative is either zero or undefined. Thus, find the derivative of the function:

\displaystyle f'(x) = \frac{d}{dx} \left[ x^a\left(2-x\right)^b\right]

By the Product Rule:

\displaystyle \begin{aligned} f'(x) &= \frac{d}{dx}\left[x^a\right] (2-x)^b + x^a\frac{d}{dx}\left[(2-x)^b\right]\\ \\ &=\left(ax^{a-1}\right)\left(2-x\right)^b + x^a\left(b(2-x)^{b-1}\cdot -1\right) \\ \\ &= x^a\left(2-x\right)^b \left[\frac{a}{x} - \frac{b}{2-x}\right] \end{aligned}

Set the derivative equal to zero and solve for <em>x: </em>

\displaystyle 0= x^a\left(2-x\right)^b \left[\frac{a}{x} - \frac{b}{2-x}\right]

By the Zero Product Property:

\displaystyle x^a (2-x)^b = 0\text{ or } \frac{a}{x} - \frac{b}{2-x} = 0

The solutions to the first equation are <em>x</em> = 0 and <em>x</em> = 2.

First, for the second equation, note that it is undefined when <em>x</em> = 0 and <em>x</em> = 2.

To solve for <em>x</em>, we can multiply both sides by the denominators.

\displaystyle\left( \frac{a}{x} - \frac{b}{2-x} \right)\left((x(2-x)\right) = 0(x(2-x))

Simplify:

\displaystyle a(2-x) - b(x) = 0

And solve for <em>x: </em>

\displaystyle \begin{aligned} 2a-ax-bx &= 0 \\ 2a &= ax+bx \\ 2a&= x(a+b) \\  \frac{2a}{a+b} &= x  \end{aligned}

So, our critical points are:

\displaystyle x = 0 , 2 , \text{ and } \frac{2a}{a+b}

We already know that f(0) = f(2) = 0.

For the third point, we can see that:

\displaystyle f\left(\frac{2a}{a+b}\right) = \left(\frac{2a}{a+b}\right)^a\left(2- \frac{2a}{a+b}\right)^b

This can be simplified to:

\displaystyle f\left(\frac{2a}{a+b}\right) = \left(\frac{2a}{a+b}\right)^a\left(\frac{2b}{a+b}\right)^b

Since <em>a</em> and <em>b</em> > 0, both factors must be positive. Thus, f(2a / (a + b)) > 0. So, this must be the maximum value.

To confirm that this is indeed a maximum, we can select values to test. Let <em>a</em> = 2 and <em>b</em> = 3. Then:

\displaystyle f'(x) = x^2(2-x)^3\left(\frac{2}{x} - \frac{3}{2-x}\right)

The critical point will be at:

\displaystyle x= \frac{2(2)}{(2)+(3)} = \frac{4}{5}=0.8

Testing <em>x</em> = 0.5 and <em>x</em> = 1 yields that:

\displaystyle f'(0.5) >0\text{ and } f'(1)

Since the derivative is positive and then negative, we can conclude that the point is indeed a maximum.

Therefore, the maximum value of f(x) occurs at:

\displaystyle x = \frac{2a}{a+b}

And is given by:

\displaystyle f_{\text{max}}(x) = \left(\frac{2a}{a+b}\right)^a\left(\frac{2b}{a+b}\right)^b

5 0
3 years ago
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