The owner of the rancho grande has 3020 yd of fencing with which to enclose a rectangular piece of grazing land situated along the straight portion of a river. if fencing is not required along the river, what are the dimensions of the largest area he can enclose?
shorter side yd
longer side yd

what is this area?
yd2

Shorter side = 755 yd

longer side   = 1510 yd  

A(max)  = 1140050 yd²

Step-by-step explanation:

The owner has 3020 yd of fencing

Lets assume:

x the shorter side of the rectangle ( we will use fencing in two sides of length x )

y the longer one

A area of the rectangle          and P the perimeter of the rectangle ( we have only three sides covered by fencing material)

We have:       A = x * y          P = 2 * x + y   ⇒  y  = P - 2*x  ⇒  y  = 3020 - 2* x

A (x) = x * ( 3020 - 2*x)  ⇒  A(x) = 3020 * x - 2* x²

Taken derivative

A´(x) = 3020 - 4 * x

If     A´(x) = 0               3020 -4*x = 0     ⇒4*x = 3020    x =  755 yd

If we take second derivative A´´(x)  = -4

so  A´´(x) < 0 so there is a maximun in point x = 755

Then

Rectangle dimensions :

x = 755 yd        ⇒  y = 3020 - 2 * x   ⇒ y = 3020 - 2 * (755)     y = 1510 yd

Maximum area is :  A(max)  = 1510 * 755  ⇒ A(max)  = 1140050 yd²

the first question mark is 6 and the second question mark is 26 (the square root)

step-by-step explanation:

answer: 204

step-by-step explanation:

we are told that the revenue is

x⋅(4080−10x)=−10x^2+4080x

this is a quadratic function, so to maximize it we should find the x-value of its vertex. recall that the x-value of the vertex of the parabola ax^2+bx+c is given by the formula −b/2a.

so in this case, we have a=−10 and b=4080, so the x-value of the vertex is

−b/2a= −(4080)/2(−10)=204