Can someone help me oit

Mathematics

1 answer:

Sergeeva-Olga [200]3 years ago

6 0

Answer:

I think its the second one

Step-by-step explanation:

You might be interested in

A square field had 3 m added to its length ad 2 m added to its width. The field then had an area of 90 m squared. Find the lengt

pogonyaev

Answer:

The length of a side of the original field = 7 m.

Step-by-step explanation:

Here, the initial field is in form of a square.

Let us assume the side of the original square field = k meters

Now, the new length of the field = ( k + 3) m

The new width of the field = ( k + 2) m

So, the new field is now a rectangle with area = 90 sq. m

AREA OF A RECTANGLE = LENGTH x WIDTH

Here, the area of the new field = New length x new width

= ( k + 3) x ( k + 2)

$90 = ( k + 3) \times ( k + 2)\\\implies k^2 + 2k + 3k + 6 = 90\\or, k^2 + 5 k - 84 = 0\\\implies k^2 + 12k - 7k -84 = 0\\or, k (k+12) -7(k+12) = 0\\\implies (k+12)(k-7) = 0$

⇒ either (k +12) = 0 ⇒ k = -12

or, ( k-7) = 0 ⇒ k = 7

But, here k = SIDE OF A FIELD, and it CANNOT be negative.

⇒ k = 7

Hence, the length of a side of the original field = 7 m.

4 0

3 years ago

Find the extreme values of f subject to both constraints

TiliK225 [7]

The Lagrangian is

$L(x,y,z,\lambda_1,\lambda_2)=z+\lambda_1(x^2+y^2-z^2)+\lambda_2(x+y+z-24)$

with partial derivatives (set equal to zero) yielding

$\begin{cases}L_x=2x\lambda_1+\lambda_2=0\\L_y=2y\lambda_1+\lambda_2=0\\L_z=1-2z\lambda_1+\lambda_2=0\\L_{\lambda_1}=x^2+y^2-z^2=0\\L_{\lambda_2}=x+y+z-24=0\end{cases}$

Subtracting the second equation from the first gives

$(2x\lambda_1+\lambda_2)-(2y\lambda_1+\lambda_2)=2\lambda_1(x-y)=0\implies x=y$

So in the fourth and fifth equations, we have

$\begin{cases}2x^2=z^2\\2x+z=24\end{cases}\implies x=24\pm12\sqrt2,z=-24\mp24\sqrt2$

There are then two critical points for $f(x,y,z)=z$ at $(24+12\sqrt2,24+12\sqrt2,-24-24\sqrt2)$ and $(24-12\sqrt2,24-12\sqrt2,-24+24\sqrt2)$ .

Clearly, $f(x,y,z)=z$ attains a local minimum at the first point of $-24-24\sqrt2$ , and a local maximum at the second point of $-24+24\sqrt2$ .