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3 years ago

For the feasibility region shown below, find the maximum value of the function P=3x+2y.

Mathematics

2 answers:

Setler [38]3 years ago

6 0

Answer:

Step-by-step explanation:

abruzzese [7]3 years ago

5 0

Given:
P = 3x + 2y

There is an accompanying graph in this problem. In the graph, there are 4 points to consider. I'll just assign letters on each point.

Point O is found in x = 0 ; y = 0 or (0,0)
Point A is found in x = 8 ; y = 0 or (8,0)
Point B is found in x = 6 ; y = 5 or (6,5)
Point C is found in x = 0 ; y = 8 or (0,8)

We will substitute x and y in the equation by its values per point.
Point A = 3(8) + 2(0) = 24 + 0 = 24
Point B = 3(6) + 2(5) = 18 + 10 = 28
Point C = 3(0) + 2(8) = 0 + 16 = 16

The maximum value of the function P = 3x+2y is 28 and its minimum value is 16.

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Suppose that the local sales tax rate is 6.25% and you purchased a used car for $16,800. What is the car’s total cost?

Tanzania [10]

Total= Car Cost + (Car Cost * Tax %)

Total= $16,800 + ($16,800 * 6.25%)
convert % to decimal form; divide % by 100

Total= $16,800 + ($16,800 * 0.0625)
multiply inside parentheses

Total= $16,800 + $1,050

Total= $17,850

ANSWER: The total cost for the used car with tax is $17,850.

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Find thd <img src="https://tex.z-dn.net/?f=%5Cfrac%7Bdy%7D%7Bdx%7D" id="TexFormula1" title="\frac{dy}{dx}" alt="\frac{dy}{dx}" a

NARA [144]

$x^3y^2+\sin(x\ln y)+e^{xy}=0$

Differentiate both sides, treating $y$ as a function of $x$ . Let's take it one term at a time.

Power, product and chain rules:

$\dfrac{\mathrm d(x^3y^2)}{\mathrm dx}=\dfrac{\mathrm d(x^3)}{\mathrm dx}y^2+x^3\dfrac{\mathrm d(y^2)}{\mathrm dx}$

$=3x^2y^2+x^3(2y)\dfrac{\mathrm dy}{\mathrm dx}$

$=3x^2y^2+6x^3y\dfrac{\mathrm dy}{\mathrm dx}$

Product and chain rules:

$\dfrac{\mathrm d(\sin(x\ln y)}{\mathrm dx}=\cos(x\ln y)\dfrac{\mathrm d(x\ln y)}{\mathrm dx}$

$=\cos(x\ln y)\left(\dfrac{\mathrm d(x)}{\mathrm dx}\ln y+x\dfrac{\mathrm d(\ln y)}{\mathrm dx}\right)$

$=\cos(x\ln y)\left(\ln y+\dfrac1y\dfrac{\mathrm dy}{\mathrm dx}\right)$

$=\cos(x\ln y)\ln y+\dfrac{\cos(x\ln y)}y\dfrac{\mathrm dy}{\mathrm dx}$

Product and chain rules:

$\dfrac{\mathrm d(e^{xy})}{\mathrm dx}=e^{xy}\dfrac{\mathrm d(xy)}{\mathrm dx}$

$=e^{xy}\left(\dfrac{\mathrm d(x)}{\mathrm dx}y+x\dfrac{\mathrm d(y)}{\mathrm dx}\right)$

$=e^{xy}\left(y+x\dfrac{\mathrm dy}{\mathrm dx}\right)$

$=ye^{xy}+xe^{xy}\dfrac{\mathrm dy}{\mathrm dx}$

The derivative of 0 is, of course, 0. So we have, upon differentiating everything,

$3x^2y^2+6x^3y\dfrac{\mathrm dy}{\mathrm dx}+\cos(x\ln y)\ln y+\dfrac{\cos(x\ln y)}y\dfrac{\mathrm dy}{\mathrm dx}+ye^{xy}+xe^{xy}\dfrac{\mathrm dy}{\mathrm dx}=0$