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Semenov [28]
3 years ago
5

1

Mathematics
1 answer:
AfilCa [17]3 years ago
6 0

Answer:

(A) Your friend will save $282.

(B) Your friend will get 23.5% off the original price of the mattress.

Step-by-step explanation:

We first need to find out the price of the mattress after the 15% discount. Our first step is to find what 15% of 1,200 is. We can do this by multiplying the two numbers.

1,200 x 15% -----> 1,200 x 0.15= 180.

15% of $1,200 is $180.

Next subtract $180 from $1,200, which equals $1,020. We can now apply the 10% off internet coupon.

1,020 x 10% ------> 1,020 x 0.10= 102.

10% of $1,020 is $102.

Next we subtract $102 from $1,020 and we get $918, the final price of the discounted mattress.

We subtract to see how much money the friend saved.

$1,200 - $918= $282.

We can get the percentage by dividing $282 by $1,200.

282 / 1200= 0.235 decimal form ----> 23.5%

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Check where the first-order partial derivatives vanish to find any critical points within the given region:

f(x,y)=2x^3+y^4\implies\begin{cases}f_x=6x^2=0\\f_y=4y^3=0\end{cases}\implies(x,y)=(0,0)

The Hessian for this function is

\mathbf H(x,y)=\begin{bmatrix}f_{xx}&f_{xy}\\f_{yx}&f_{yy}\end{bmatrix}=\begin{bmatrix}12x&0\\0&12y^2\end{bmatrix}

with \det\mathbf H(0,0)=0, so unfortunately the second partial derivative test fails. However, if we take x=0 we see that f(x,y)>0 for different values of y; if we take y=0 we see f(x,y) takes on both positive and negative values. This indicates (0, 0) is neither the site of an extremum nor a saddle point.

Now check for points along the boundary. We can parameterize the boundary by

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F(t)=f(8\cos t,8\sin t)=2^{10}\cos^3t+2^{12}\sin^4t

\implies F'(t)=3\cdot2^{10}\cos^2t(-\sin t)+2^{14}\sin^3t\cos t=2^{10}\sin t\cos t(16\sin^2t-3\cos t)

F'(t)=0\implies\begin{matrix}\sin t=0\implies t=0,\,t=\pi\\\\\cos t=0\implies t=\dfrac\pi2,\,t=\dfrac{3\pi}2\\\\16\sin^2t-3\cos t=0\implies t=2\tan^{-1}\sqrt{\dfrac{\sqrt{1033}-32}3},\,t=2\pi-2\tan^{-1}\sqrt{\dfrac{\sqrt{1033}-32}3}\end{matrix}

At these critical points, we get

F(0)=1024

F\left(2\tan^{-1}\sqrt{\dfrac{\sqrt{1033}-32}3}\right)\approx893

F\left(\dfrac\pi2\right)=4096

F(\pi)=-1024

F\left(\dfrac{3\pi}2\right)=4096

F\left(2\pi-2\tan^{-1}\sqrt{\dfrac{\sqrt{1033}-32}3}\right)\approx893

We only care about 3 of these results.

t=\dfrac\pi2\implies(x,y)=(0,8)

t=\pi\implies(x,y)=(-8,0)

t=\dfrac{3\pi}2\implies(x,y)=(0,-8)

So to recap, we found that f(x,y) attains

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