Help would appreciated

Find the minimum value of the function for the polygonal convex set determined by the given system of inequalities.

lutik1710 [3]

Answer:

Option b (4,1)

Step-by-step explanation:

The region given by the system of inequalities is shown in the graph. We must look within this region for the point that minimizes the objective function $f(x, y) = 8x + 8y$

The minimum points are found in the lower vertices of the region.

These vertices are found by equating the equations of the lines::

$3x+2y=14\\-5x +5y=10$

-------------------

$x = 2\\y = 4$

$-8x + 3y = -29\\3x + 2y = 14$

---------------------

$x = 4\\y = 1$

The lower vertices are:

(4, 1) (2, 4)

Now we substitute both points in the objective function to see which of them we get the lowest value of $f(x, y)$

$f(4, 1) = 8(4) +8(1) = 40\\f(2, 4) = 8(2) + 8(4) = 48$

Then the value that minimizes f(x, y) is (4,1).

Option b

5 0

3 years ago

Y=x.arctan(x)^1/2. find dy/dx. pls show steps

Whitepunk [10]

$y=x(\arctan x)^{1/2}$

Use the product rule first:

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\mathrm dx}{\mathrm dx}(\arctan x)^{1/2}+x\dfrac{\mathrm d(\arctan x)^{1/2}}{\mathrm dx}$

$\dfrac{\mathrm dy}{\mathrm dx}=(\arctan x)^{1/2}+x\dfrac{\mathrm d(\arctan x)^{1/2}}{\mathrm dx}$

Use the chain rule to compute the derivative of $(\arctan x)^{1/2}$ . Let $z=(\arctan x)^{1/2}$ and take $u=\arctan x$ , so that by the chain rule

$\dfrac{\mathrm dz}{\mathrm dx}=\dfrac{\mathrm dz}{\mathrm du}\dfrac{\mathrm du}{\mathrm dx}$

$\dfrac{\mathrm dz}{\mathrm du}=\dfrac{\mathrm du^{1/2}}{\mathrm du}=\dfrac12u^{-1/2}$

$\dfrac{\mathrm du}{\mathrm dx}=\dfrac{\mathrm d\arctan x}{\mathrm dx}=\dfrac1{1+x^2}$

$\implies\dfrac{\mathrm d(\arctan x)^{1/2}}{\mathrm dx}=\dfrac1{2(\arctan x)^{1/2}(1+x^2)}$

So we have

$\dfrac{\mathrm dy}{\mathrm dx}=(\arctan x)^{1/2}+\dfrac x{2(\arctan x)^{1/2}(1+x^2)}$

You can rewrite this a bit by factoring $(\arctan x)^{-1/2}$ , just to make it look neater:

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac1{2(\arctan x)^{1/2}}\left(2\arctan x+\dfrac x{1+x^2}\right)$

3 0

3 years ago

The length of a rectangular floor is two ft more than its width. If the area of the floor is 63 ft, find the length and the widt

NemiM [27]

Answer:

width = 7 and length = 9

Step-by-step explanation:

Let's assume width = w, and length = 2 + w

So area of a rectangle = width * length

63 ft² = w (2 + w)

63 = 2w + w²

w² + 2w - 63 = 0

Use the quadratic formula to solve this

here, a = 1, b = 2 and c = -63

w = 7 or -9

The dimensions of a rectangle cannot be negative. So we take the dimension, w = 7.

Width = 7 and Length = 7+2 = 9

3 0

2 years ago

What is the slope of the equation that is represented by the table below?

yuradex [85]

Answer:A

Step-by-step explanation:

You use the equation:

m= (y2-y1) ÷ ( x2-x1)

m= (12-9) ÷ (2-1)

m= 3÷1

m= 3

3 0

4 years ago

Read 2 more answers

I am really confused can someone please help?!

pshichka [43]

If the triangle is translated left 9, this shows that all the x values will decrease by nine, becoming, -15, -5, and -8.

If the triangle is translated up 12, this shows that all the y values will increase by twelve, becoming, 13, 12, and 15.

Thus, the new co-ordinates are <u>X(-15,13), Y(-5,12), Z(-8,15).</u>

5 0

3 years ago