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4 years ago

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Find two numbers whose difference is 124 and whose product is a minimum. (smaller number) (larger number)

2 answers:

icang [17]4 years ago

8 0

1 and 125

$125 - 1 = 124$
$125 \times 1 = 125$
good luck

Galina-37 [17]4 years ago

3 0

For this case, the first thing we must do is define variables.

We have then:

x: unknown number 1

y: unknown number 2

The difference of both numbers is 124:

$x - y = 124$

The product of both numbers is:

$P = x * y$

Writing the product based on a variable we have:

$P (x) = x * (x-124)$

Rewriting:

$P (x) = (x ^ 2-124x)$

To find the minimum of the function, we must derive:

$P '(x) = 2x-124$

We equal zero and clear x:

$2x-124 = 0\\2x = 124\\x = 124/2\\x = 62$

Then, the value of the other number is given by:

$y = x-124\\y = 62-124\\y = -62$

Answer:

The searched numbers that meet both conditions are:

smaller number = 62

larger number = -62

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The surface area of a cylinder is increasing at a rate of 9 pi square meters per hour. The height of the cylinder is fixed at 3

Alekssandra [29.7K]

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Step-by-step explanation:

Since, the surface area of a cylinder,

$A= 2\pi r^2 + 2\pi rh$ ................(1)

Where,

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h = height,

If $A= 36\pi\text{ square meters}, h = 3\text{ meters}$

$36\pi = 2\pi r^2 + 2\pi r(3)$

$18 = r^2 + 3r$

$\implies r^2 + 3r - 18=0$

$r^2 + 6r - 3r - 18 = 0$ ( by middle term splitting )

$r(r+6)-3(r+6)=0$

$(r-3)(r+6)=0$

By zero product property,

r = 3 or r = - 6 ( not possible )

Thus, radius, r = 3 meters,

Now, differentiating equation (1) with respect to t ( time ),

$\frac{dA}{dt}= 4\pi r\frac{dr}{dt} +2\pi(r\frac{dh}{dt} + h\frac{dr}{dt})$

∵ h = constant, ⇒ dh/dt = 0,

$\frac{dA}{dt} = 4\pi r \frac{dr}{dt} +2\pi h \frac{dr}{dt}$

We have, $\frac{dA}{dt}=9\pi\text{ square meters per hour}, r = h = 3\text{ meters}$

$9\pi = 4\pi (3) \frac{dr}{dt}+2\pi (3)\frac{dr}{dt}$

$9\pi = (12\pi + 6\pi )\frac{dr}{dt}$

$9\pi = 18\pi \frac{dr}{dt}$

$\implies \frac{dr}{dt} =\frac{1}{2}\text{ meter per hour}$

Now,

Volume of a cylinder,

$V=\pi r^2 h$

Differentiating w. r. t. t,

$\frac{dV}{dt}=\pi ( r^2 \frac{dh}{dt}+h(2r)\frac{dr}{dt})=\pi ((3)(6) (\frac{1}{2})) = 9\pi \text{ cubic meters per hour}$

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3 years ago

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