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dedylja [7]
2 years ago
15

A rectangle is to be inscribed in an isosceles right triangle in such a way that one vertex of the rectangle is the intersection

point of the legs of the triangle and the opposite vertex lies on the hypotenuse. Find the largest area (in cm 2 ) of the rectangle and its dimensions (in cm) given that the two equal legs of the triangle have length 4.
Mathematics
1 answer:
svet-max [94.6K]2 years ago
5 0

Answer:

x  =  2  cm

y  = 2  cm

A(max) =  4 cm²

Step-by-step explanation: See Annex

The right isosceles triangle has two 45° angles and the right angle.

tan 45°  =  1  =  x / 4 - y        or     x  =  4  -  y     y  =  4  -  x

A(r)  =  x* y

Area of the rectangle as a function of x

A(x)  =  x  *  (  4  -  x )       A(x)  =  4*x  -  x²

Tacking derivatives on both sides of the equation:

A´(x)  =  4 - 2*x             A´(x)  =  0            4   -  2*x  =  0

2*x  =  4

x  =  2  cm

And  y  =  4  - 2  =  2  cm

The rectangle of maximum area result to be a square of side 2 cm

A(max)  = 2*2  =  4 cm²

To find out if A(x) has a maximum in the point  x  =  2

We get the second derivative

A´´(x)  =  -2           A´´(x)  <  0   then A(x) has a maximum at  x = 2

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If 5/8ths is equal to $531.25 what is the remaining portion of the number.
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An investment of $8,000 earns interest at an annual rate of 7% compounded continuously. Complete parts (A) and (B) below. Click
Sergeeva-Olga [200]

Answer:

A.    \mathtt{\dfrac{dA}{dt}|_{t=2}=644.15}

B.    \mathtt{\dfrac{dA}{dt}|_{t = 5.79}= 839.86 }

Step-by-step explanation:

Given that:

An investment of  Amount = $8000

earns  at an annual rate of interest = 7% = 0.07 compounded continuously

The objective is to :

A)  Find the instantaneous rate of change of the amount in the account after 2 year(s).

we all know that:

A = Pe^{rt}

where;

A = (8000) \ e ^{0.7t}

The instantaneous rate of change = \dfrac{dA}{dt}

\dfrac{dA}{dt} = \dfrac{d}{dt}(8000 \ e ^{0.07t} )

= 8000 \dfrac{d}{dt}e^{0.07 \ t}

\dfrac{dA}{dt}= 8000 (0.07)e^{0.07 \ t}

\dfrac{dA}{dt}= 560 e^{0.07 \ t}

At t = 2 years; the instantaneous rate of change is:

\dfrac{dA}{dt}|_{t=2}= 560 e^{0.07 \times 2}

\mathtt{\dfrac{dA}{dt}|_{t=2}=644.15}

(B) Find the instantaneous rate of change of the amount in the account at the time the amount is equal to $12,000.

Here the amount = 12000

12000 = (8000)e^{0.07 \ t}

\dfrac{12000 }{8000}= e^{0.07 \ t}

1.5= e^{0.07 \ t}

㏑(1.5) = 0.07 t

0.405465 = 0.07 t

t = 0.405465 /0.07

t = 5.79

\dfrac{dA}{dt}= 560 e^{0.07 \ t}

At t = 5.79

\dfrac{dA}{dt}|_{t = 5.79}= 560 e^{0.07 \times 5.79}

\mathtt{\dfrac{dA}{dt}|_{t = 5.79}= 839.86 }

7 0
3 years ago
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