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a_sh-v [17]
1 year ago
15

I would like to create a rectangular vegetable patch. The fencing for the east and west sides costs $4 per foot, and the fencing

for the north and south sides costs only $2 per foot. I have a budget of $176 for the project. What are the dimensions of the vegetable patch with the largest area I can enclose?
Mathematics
1 answer:
ss7ja [257]1 year ago
5 0
X = E/W dimension 
<span>y = N/S dimension </span>

<span>4x + 4x + 2y + 2y = 64 </span>
<span>8x + 4y = 64 </span>
<span>4y = 64 - 8x </span>
<span>y = 16 - 2x </span>

<span>Area = xy = x(16 - 2x) = 16x - 2x^2 </span>

<span>Maximum of y = ax^2 + bx + c is when x = -b / 2a </span>

<span>so x = -16 / -4 = 4 </span>
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Timothy makes reduced copies of a photograph that has an actual length of 8 in. Each time he presses the reduce button on the co
Aleksandr-060686 [28]

Answer:

length of the photograph will be 4.2 in. after pressing the button 5 times.

Step-by-step explanation:

By pressing the button, every time size of the photograph gets reduced by 12%.

Therefore, the sequence formed by the reduced sizes of the photo will be a geometric sequence and the formula for the size of the reduced image will be,

L = l(1-\frac{12}{100})^{n}

Where l = Actual length of the photograph

L = length of the reduced image

n = Number of times the button has been pressed

For l = 8 in. and n = 5

L = 8(1-0.12)^{5}

  = 8(0.88)^{5}

  = 4.22 in

L ≈ 4.2 in.

Therefore, length of the photograph will be 4.2 in. after pressing the button 5 times.

3 0
2 years ago
Find the value of $x$ that maximizes $$f(x) = \log (-20x + 12\sqrt{x}).$$ If there is no maximum value, write "NONE".
kakasveta [241]
The function is written as:
f(x) = log(-20x + 12√x)
To find the maximum value, differentiate the equation in terms of x, then equate it to zero. The solution is as follows.

The formula for differentiation would be:
d(log u)/dx = du/u ln(10)
Thus,
d/dx = (-20 + 6/√x)/(-20x + 12√x)(ln 10) = 0
-20 + 6/√x = 0
6/√x = 20
x = (6/20)² = 9/100

Thus,
f(x) =  log(-20(9/100)+ 12√(9/100)) = 0.2553

<em>The maximum value of the function is 0.2553.</em>
5 0
2 years ago
The random variable X is normally distributed with mean 82 and standard deviation 7.4. Find the value of q such that P(82 − q &l
mezya [45]
P(82 - q < x < 82 + q) = 0.44
P(x < 82 + q) - P(82 - q) = 0.44
P(z < (82 + q - 82)/7.4 - P(z < (82 - q - 82)/7.4) = 0.44
P(z < q/7.4) - P(z < -q/7.4) = 0.44
P(z < q/7.4) - (1 - P(z < q/7.4) = 0.44
P(z < q/7.4) - 1 + P(z < q/7.4) = 0.44
2P(z < q/7.4) - 1 = 0.44
2P(z < q/7.4) = 1.44
P(z < q/7.4) = 0.72
P(z < q/7.4) = P(z < 0.583)
q/7.4 = 0.583
q = 0.583 x 7.4 = 4.31
8 0
1 year ago
The telephone company is planning to introduce two new types of executive communications systems that it hopes to sell to its la
cluponka [151]

Answer:

x = 31 hundred dolars   and      

y = 91/2 = 45.5 hundred dolars

Step-by-step explanation:

Given

R(x) = (40−8x+5y)*x + (50+9x−7y)*y

C(x) = (40−8x+5y)*10 + (50+9x−7y)*29

We can use the equation

P(x) = R(x) - C(x)

where

P(x) is the profit

R(x) is the revenue

and C(x) is the costs

In order to maximize the telephone company's profit, we apply

P'(x) = R(x)' - C(x)' = 0

⇒ R(x)' = ((40−8x+5y)*x + (50+9x−7y)*y)' = (40x-8x²+14xy+50y-7y²)'

⇒ C(x)' = ((40−8x+5y)*10 + (50+9x−7y)*29)' = (1850+181x-153y)'

⇒ P'(x) = -8x²-7y²-141x+203y+14xy-1850

The first-order partial derivatives of these functions are

Px(x,y) = -16x-141+14y

Py(x,y) = -14y+203+14x

Setting these equal to zero and solving we obtain:

-16x+14y-141 = 0

14x-14y+203=0

we get the solution

x = 31     and       y = 91/2 = 45.5

Finally, the company should produce  3100  units of the first system, and  4550 units of the second system.

8 0
2 years ago
9
Alexeev081 [22]

<em>so</em><em> </em><em>the</em><em> </em><em>answer</em><em> </em><em>is</em><em> </em><em>3</em><em>6</em><em>.</em>

<em>Hope</em><em> </em><em>this</em><em> </em><em>will</em><em> </em><em>help</em><em> </em><em>u</em><em>.</em><em>.</em><em>.</em>

7 0
1 year ago
Read 2 more answers
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