Help Optimization Problems!?

Question

Calculus is kicking my butt. Can anyone help me start off these two problems. I'm not looking for the answer just a start off. Thank You! 1.) A rectangular storage container with an open top is to have a volume of 10m^3. The length of its base is twice the width. Material for the base costs $10 per square meter. Material for the sides costs $6 per square meter. Find the cost of materials for the cheapest such container. 2.) Find the area for the largest rectangle that can be inscribed in a right triangle with the legs 3 cm and 4 cm if two sides of the rectangle lie along the legs. THANK YOU SO MUCH AGAIN :)

Answer 1

fixed volume means our constraint eqn is V=l*w*h
10 = (2w)w(h) because l=2w
2 w^2 h=10 ===> h = 5/w^2

MAIN equation to minimize cost. Find cost eqn, then set C'=0
cost.......base....+....2sideslh + 2sides wh
Cost = 10*w*2w + 2 * 6 *h*2w + 2*6 *wh
C = 20w^2 +24hw + 12hw
C = 20w^2 +36hw===> to get to single variable plug in h=5/w^2 from above
C = 20w^2 +36(5/w^2)w
C = 20w^2 + 180 / w
now find C' and set equal to zero


#2 Place triangle on axes, so 3 on y-axis, 4 on x-axis, in QI
So the equation of the hypotenuse is y = -3/4 x +3
So now we know any point on the line is (x , -3/4x + 3)
Point must be on line
Area = base*height
A = x*y
but we know y = -3/4x + 3
A = x(-3/4x+3)
simplify before A'=0


:)

FYI to check your work answers
#1 w = 1.651 and Cost is $163.54
#2 x=2 ===> A=3 square cm

Answer 2

1.) A rectangular storage container with an open top is to have a volume of 10m^3. The length of its base is twice the width. Material for the base costs $10 per square meter. Material for the sides costs $6 per square meter. Find the cost of materials for the cheapest such container.

Let
2x = length base
x = width base
h = height box
V = volume box
S = surface area box
C = cost to make box

We have

V = (2x)xh = 2x²h = 1000
h = 1000/(2x²) = 500/x²

S = (2x)x + 2xh + 2(2x)h = 2x² + 4xh
C = 10*2x² + 6*4xh = 20x² + 24xh

Substitute the value for h into the cost function and take the derivative of C with respect to x.

There's your start off.
______________________

2.) Find the area for the largest rectangle that can be inscribed in a right triangle with the legs 3 cm and 4 cm if two sides of the rectangle lie along the legs.

Let
h = height rectangle
w = width rectangle
A = area rectangle

We have

A = wh

3w + 4h = 12

Solve for either w or h and plug back into the formula for A. Suppose you solve for h. Then after plugging back in take the derivative dA/dw.

Answer 3

The equation of a line is y = mx + b meaning at (6, 10), the equation is 10 = m(6) + b, or 10 = 6m + b or b = 10 - 6m The x-intercept of this line represents the bottom of the triangle. The x-intercept is discovered by technique of creating y = 0, so y = mx + b 0 = mx + b -b = mx x = -b/m <==== the bottom. The y-intercept represents the proper of the triangle, discovered by technique of creating x = 0. y = mx + b y = b <====== the proper The formula of the realm of the triangle is (a million/2)(base)(proper). for this reason, A = (a million/2)(-b/m)(b) A = (a million/2)(-b^2 / m) A = (-a million/2)(b^2/m) yet all of us keep in mind that b = 10 - 6m, so A = (-a million/2)( [10 - 6m]^2 / m ) So this may be our section formula, A(m) A(m) = (-a million/2)( [10 - 6m]^2 / m ) we ought to reduce the realm, so we take the by-product as general. A'(m) = (-a million/2) [ 2(10 - 6m)(-6)m - (10 - 6m)^2 (a million) ] / [ m^2 ] A'(m) = (-a million/2) [ (-12m)(10 - 6m) - (10 - 6m)^2 ] / [ m^2 ] A'(m) = (-a million/2) [ (10 - 6m)(-12m - (10 - 6m)) ] / [ m^2 ] A'(m) = (-a million/2) [ (10 - 6m)(-12m - 10 + 6m) ] / [ m^2 ] A'(m) = (-a million/2) [ (10 - 6m)(-6m - 10) ] / [ m^2 ] Now, we make A'(m) = 0. 0 = (-a million/2) [ (10 - 6m)(-6m - 10) ] / [ m^2 ] severe values: 5/3, -5/3 we may be able to reject 5/3 because it really is a marvelous slope (the in elementary words thanks to type triangles from the first quadrant is that if the slope is adverse). That leaves -5/3, so this tells us the minimum section takes position at the same time as the slope m = -5/3. to discover the equation of the line, all we ought to do is stumble upon severe college the thanks to discover the equation of the line with slope m = = -5/3 and through (6, 10). (y2 - y1)/(x2 - x1) = -5/3 (y - 10)/(x - 6) = -5/3 y - 10 = (-5/3)(x - 6) y - 10 = (-5/3)x + 10 y = (-5/3)x + 20

Answer 4

Optimization problems are a hassle. Usually, you have to set up your function to be optimized based on one variable, but the variable is related to others in another formula. Like this one. Let x be the width in meters, 2x be the length, and h be the height. They are related by
Volume = x * 2x * h or 10 = 2 x^2 h.
You want to optimize material costs. Let that be C
C = 10 * 2 x^2 + 6 (2[ x * h]+ 2 [2x * h])
C= 20 x^2 + 12 (3x * h)
So substiute in 5/x^2 for h, take the derivitive, set it equal to 0, and solve for x and h, and finally C.

Second problem. Let the height of the rectangle be h and the base b. They are linked through the triangle such that h^2+(4-b)^2 + b^2+(3-h)^2 =25
Your are trying to maximize bh. Again, you have to solve the above for h or b, substiute the result in A= bh, and for the dimension selected, take the derivitive, set it equal to zero, and solve.

Answer 5

1.) Start by drawing the container and listing what you know about it.

We know that V = (L)(W)(H). We already know the value of V, and we also already know L and W in terms of x. Use that to find H in terms of x.

Once you get the three dimensions of the rectangular container in terms of x, you can write an equation to calculate the area of each side and base of the container also in terms of x. Add up the sides and multiply by $6 plus the base multiplied by $10 to get the cost equation. Derive and find the minimum. Plug x back in to find the actual cost.